Split (1)

train · 20k rows

problem stringlengths 15 4.7k	solution stringlengths 2 11.9k	answer stringclasses 51 values	problem_type stringclasses 8 values	question_type stringclasses 4 values	problem_is_valid stringclasses 1 value	solution_is_valid stringclasses 1 value	source stringclasses 6 values	synthetic bool 1 class
After a cyclist has gone $ \frac{2}{3}$ of his route, he gets a flat tire. Finishing on foot, he spends twice as long walking as he did riding. How many times as fast does he ride as walk?	1. Let the total distance of the route be $ D $. 2. The cyclist rides $ \frac{2}{3}D $ of the route and walks the remaining $ \frac{1}{3}D $. 3. Let $ t_r $ be the time spent riding and $ t_w $ be the time spent walking. 4. According to the problem, the time spent walking is twice the time spent riding: \[ t_w = 2t_r \] 5. Let $ v_r $ be the riding speed and $ v_w $ be the walking speed. 6. Using the relationship between distance, speed, and time, we have: \[ \frac{2}{3}D = v_r t_r \quad \text{(1)} \] \[ \frac{1}{3}D = v_w t_w \quad \text{(2)} \] 7. Substitute $ t_w = 2t_r $ into equation (2): \[ \frac{1}{3}D = v_w (2t_r) \] \[ \frac{1}{3}D = 2v_w t_r \quad \text{(3)} \] 8. From equation (1), solve for $ t_r $: \[ t_r = \frac{\frac{2}{3}D}{v_r} = \frac{2D}{3v_r} \] 9. Substitute $ t_r $ from above into equation (3): \[ \frac{1}{3}D = 2v_w \left(\frac{2D}{3v_r}\right) \] \[ \frac{1}{3}D = \frac{4D v_w}{3v_r} \] 10. Cancel $ D $ from both sides: \[ \frac{1}{3} = \frac{4 v_w}{3 v_r} \] 11. Multiply both sides by $ 3v_r $: \[ v_r = 4 v_w \] 12. Therefore, the cyclist rides 4 times as fast as he walks. The final answer is $ \boxed{4} $.	4	Algebra	math-word-problem	Yes	Yes	aops_forum	false
After a typist has written ten letters and had addressed the ten corresponding envelopes, a careless mailing clerk inserted the letters in the envelopes at random, one letter per envelope. What is the probability that [b]exactly[/b] nine letters were inserted in the proper envelopes?	1. Let's denote the letters as $ L_1, L_2, \ldots, L_{10} $ and the corresponding envelopes as $ E_1, E_2, \ldots, E_{10} $. 2. We are asked to find the probability that exactly nine letters are inserted into their corresponding envelopes. 3. If nine letters are correctly placed in their corresponding envelopes, then the tenth letter must also be correctly placed in its corresponding envelope because there is only one envelope left. 4. This means that if nine letters are correctly placed, the tenth letter must also be correctly placed, resulting in all ten letters being correctly placed. 5. Therefore, it is impossible to have exactly nine letters correctly placed without the tenth letter also being correctly placed. 6. Hence, the probability of exactly nine letters being correctly placed is zero. \[ \boxed{0} \]	0	Logic and Puzzles	math-word-problem	Yes	Yes	aops_forum	false
Car A is traveling at 20 miles per hour. Car B is 1 mile behind, following at 30 miles per hour. A fast fly can move at 40 miles per hour. The fly begins on the front bumper of car B, and flies back and forth between the two cars. How many miles will the fly travel before it is crushed in the collision?	1. Determine the time it takes for Car B to catch up to Car A: - Car A is traveling at 20 miles per hour. - Car B is traveling at 30 miles per hour and is 1 mile behind Car A. - The relative speed of Car B with respect to Car A is $30 - 20 = 10$ miles per hour. - The time $t$ it takes for Car B to catch up to Car A is given by: \[ t = \frac{\text{distance}}{\text{relative speed}} = \frac{1 \text{ mile}}{10 \text{ miles per hour}} = \frac{1}{10} \text{ hours} \] 2. Calculate the distance the fly travels: - The fly moves at a constant speed of 40 miles per hour. - The time the fly has to travel before the collision is $\frac{1}{10}$ hours. - The distance $d$ the fly travels is given by: \[ d = \text{speed} \times \text{time} = 40 \text{ miles per hour} \times \frac{1}{10} \text{ hours} = 4 \text{ miles} \] Conclusion: \[ \boxed{4} \]	4	Calculus	math-word-problem	Yes	Yes	aops_forum	false
Given a random string of 33 bits (0 or 1), how many (they can overlap) occurrences of two consecutive 0's would you expect? (i.e. "100101" has 1 occurrence, "0001" has 2 occurrences)	1. Determine the probability of two consecutive bits being both 0: - Each bit in the string can be either 0 or 1 with equal probability, i.e., $ \frac{1}{2} $. - The probability that two consecutive bits are both 0 is: \[ \left( \frac{1}{2} \right)^2 = \frac{1}{4} \] 2. Calculate the number of possible locations for a two-bit string in a 33-bit string: - A 33-bit string has 32 possible locations for a two-bit substring (since each pair of consecutive bits can start at any of the first 32 positions). - This is because the first bit of the two-bit substring can be at any position from 1 to 32. 3. Calculate the expected number of occurrences of two consecutive 0's: - The expected number of occurrences is the product of the number of possible locations and the probability of each location having two consecutive 0's. - Therefore, the expected number of occurrences is: \[ 32 \times \frac{1}{4} = 8 \] The final answer is $\boxed{8}$.	8	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
There exist two positive numbers $ x$ such that $ \sin(\arccos(\tan(\arcsin x)))\equal{}x$. Find the product of the two possible $ x$.	1. Let $\arcsin x = \theta$. Then, $x = \sin \theta$ and $\theta \in [0, \frac{\pi}{2}]$ since $x$ is a positive number. 2. We need to find $\tan(\arcsin x)$. Using the right triangle representation, we have: \[ \sin \theta = x \quad \text{and} \quad \cos \theta = \sqrt{1 - x^2} \] Therefore, \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x}{\sqrt{1 - x^2}} \] 3. Let $\beta = \arccos \left( \frac{x}{\sqrt{1 - x^2}} \right)$. Then, $\cos \beta = \frac{x}{\sqrt{1 - x^2}}$. 4. We need to find $\sin \beta$. Using the right triangle representation again, we have: \[ \cos \beta = \frac{x}{\sqrt{1 - x^2}} \quad \text{and} \quad \sin \beta = \sqrt{1 - \cos^2 \beta} \] Therefore, \[ \sin \beta = \sqrt{1 - \left( \frac{x}{\sqrt{1 - x^2}} \right)^2} = \sqrt{1 - \frac{x^2}{1 - x^2}} = \sqrt{\frac{1 - x^2 - x^2}{1 - x^2}} = \sqrt{\frac{1 - 2x^2}{1 - x^2}} \] 5. Given that $\sin(\arccos(\tan(\arcsin x))) = x$, we have: \[ \sqrt{\frac{1 - 2x^2}{1 - x^2}} = x \] 6. Squaring both sides, we get: \[ \left( \sqrt{\frac{1 - 2x^2}{1 - x^2}} \right)^2 = x^2 \implies \frac{1 - 2x^2}{1 - x^2} = x^2 \] 7. Multiplying both sides by $1 - x^2$, we obtain: \[ 1 - 2x^2 = x^2 (1 - x^2) \implies 1 - 2x^2 = x^2 - x^4 \] 8. Rearranging terms, we get: \[ x^4 - 3x^2 + 1 = 0 \] 9. Let $y = x^2$. Then, the equation becomes: \[ y^2 - 3y + 1 = 0 \] 10. Solving this quadratic equation using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get: \[ y = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \] 11. Therefore, the solutions for $x^2$ are: \[ x^2 = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad x^2 = \frac{3 - \sqrt{5}}{2} \] 12. Since $x$ is positive, we take the positive square roots: \[ x = \sqrt{\frac{3 + \sqrt{5}}{2}} \quad \text{and} \quad x = \sqrt{\frac{3 - \sqrt{5}}{2}} \] 13. We are asked to find the product of the two possible values of $x$. Therefore, we calculate: \[ \left( \sqrt{\frac{3 + \sqrt{5}}{2}} \right) \left( \sqrt{\frac{3 - \sqrt{5}}{2}} \right) = \sqrt{\left( \frac{3 + \sqrt{5}}{2} \right) \left( \frac{3 - \sqrt{5}}{2} \right)} \] 14. Simplifying the product inside the square root: \[ \left( \frac{3 + \sqrt{5}}{2} \right) \left( \frac{3 - \sqrt{5}}{2} \right) = \frac{(3 + \sqrt{5})(3 - \sqrt{5})}{4} = \frac{9 - 5}{4} = \frac{4}{4} = 1 \] 15. Therefore, the product of the two possible values of $x$ is: \[ \sqrt{1} = 1 \] The final answer is $\boxed{1}$	1	Other	math-word-problem	Yes	Yes	aops_forum	false
There are three bins: one with 30 apples, one with 30 oranges, and one with 15 of each. Each is labeled "apples," "oranges," or "mixed." Given that all three labels are wrong, how many pieces of fruit must you look at to determine the correct labels?	1. Identify the problem constraints: - There are three bins, each labeled incorrectly. - One bin contains only apples, one contains only oranges, and one contains a mix of both. - The labels are "apples," "oranges," and "mixed." 2. Pick a fruit from the bin labeled "mixed": - Since the labels are all wrong, the bin labeled "mixed" cannot actually contain a mix of apples and oranges. It must contain either only apples or only oranges. 3. Determine the content of the bin labeled "mixed": - Suppose you pick a fruit from the bin labeled "mixed" and it is an apple. This means the bin labeled "mixed" actually contains only apples. 4. Relabel the bins based on the first observation: - Since the bin labeled "mixed" is actually the apple bin, the bin labeled "oranges" cannot contain only oranges (because all labels are wrong). Therefore, it must be the mixed bin. - The remaining bin, which is labeled "apples," must then be the orange bin. 5. Verify the logic: - If the fruit picked from the bin labeled "mixed" is an apple: - The bin labeled "mixed" is actually the apple bin. - The bin labeled "oranges" is the mixed bin. - The bin labeled "apples" is the orange bin. - If the fruit picked from the bin labeled "mixed" is an orange: - The bin labeled "mixed" is actually the orange bin. - The bin labeled "apples" is the mixed bin. - The bin labeled "oranges" is the apple bin. 6. Conclusion: - By picking just one fruit from the bin labeled "mixed," you can determine the correct labels for all three bins. The final answer is $\boxed{1}$.	1	Logic and Puzzles	math-word-problem	Yes	Yes	aops_forum	false
Peter Pan and Crocodile are each getting hired for a job. Peter wants to get paid 6.4 dollars daily, but Crocodile demands to be paid 10 cents on day 1, 20 cents on day 2, 40 cents on day 3, 80 cents on day 4, and so on. After how many whole days will Crocodile's total earnings exceed that of Peter's?	To determine after how many whole days Crocodile's total earnings will exceed Peter's, we need to set up the equations for their earnings and solve for the number of days, $ n $. 1. Calculate Peter's total earnings after $ n $ days: Peter earns $6.4 per day. Therefore, after $ n $ days, Peter's total earnings, $ P(n) $, is: \[ P(n) = 6.4n \] 2. Calculate Crocodile's total earnings after $ n $ days: Crocodile's earnings double each day, starting from 10 cents on day 1. This forms a geometric series where the first term $ a = 0.1 $ dollars and the common ratio $ r = 2 $. The total earnings after $ n $ days, $ C(n) $, is the sum of the first $ n $ terms of this geometric series: \[ C(n) = 0.1 + 0.2 + 0.4 + \cdots + 0.1 \cdot 2^{n-1} \] The sum of the first $ n $ terms of a geometric series is given by: \[ S_n = a \frac{r^n - 1}{r - 1} \] Substituting $ a = 0.1 $ and $ r = 2 $: \[ C(n) = 0.1 \frac{2^n - 1}{2 - 1} = 0.1 (2^n - 1) \] 3. Set up the inequality to find when Crocodile's earnings exceed Peter's: We need to find the smallest $ n $ such that: \[ C(n) > P(n) \] Substituting the expressions for $ C(n) $ and $ P(n) $: \[ 0.1 (2^n - 1) > 6.4n \] Simplifying: \[ 2^n - 1 > 64n \] \[ 2^n > 64n + 1 \] 4. Solve the inequality: We need to find the smallest integer $ n $ for which $ 2^n > 64n + 1 $. We can test values of $ n $ to find the smallest solution: - For $ n = 1 $: \[ 2^1 = 2 \quad \text{and} \quad 64 \cdot 1 + 1 = 65 \quad \Rightarrow \quad 2 < 65 \] - For $ n = 2 $: \[ 2^2 = 4 \quad \text{and} \quad 64 \cdot 2 + 1 = 129 \quad \Rightarrow \quad 4 < 129 \] - For $ n = 3 $: \[ 2^3 = 8 \quad \text{and} \quad 64 \cdot 3 + 1 = 193 \quad \Rightarrow \quad 8 < 193 \] - For $ n = 4 $: \[ 2^4 = 16 \quad \text{and} \quad 64 \cdot 4 + 1 = 257 \quad \Rightarrow \quad 16 < 257 \] - For $ n = 5 $: \[ 2^5 = 32 \quad \text{and} \quad 64 \cdot 5 + 1 = 321 \quad \Rightarrow \quad 32 < 321 \] - For $ n = 6 $: \[ 2^6 = 64 \quad \text{and} \quad 64 \cdot 6 + 1 = 385 \quad \Rightarrow \quad 64 < 385 \] - For $ n = 7 $: \[ 2^7 = 128 \quad \text{and} \quad 64 \cdot 7 + 1 = 449 \quad \Rightarrow \quad 128 < 449 \] - For $ n = 8 $: \[ 2^8 = 256 \quad \text{and} \quad 64 \cdot 8 + 1 = 513 \quad \Rightarrow \quad 256 < 513 \] - For $ n = 9 $: \[ 2^9 = 512 \quad \text{and} \quad 64 \cdot 9 + 1 = 577 \quad \Rightarrow \quad 512 < 577 \] - For $ n = 10 $: \[ 2^{10} = 1024 \quad \text{and} \quad 64 \cdot 10 + 1 = 641 \quad \Rightarrow \quad 1024 > 641 \] Therefore, the smallest $ n $ for which $ 2^n > 64n + 1 $ is $ n = 10 $. The final answer is $ \boxed{10} $	10	Inequalities	math-word-problem	Yes	Yes	aops_forum	false
A rope of length 10 [i]m[/i] is tied tautly from the top of a flagpole to the ground 6 [i]m[/i] away from the base of the pole. An ant crawls up the rope and its shadow moves at a rate of 30 [i]cm/min[/i]. How many meters above the ground is the ant after 5 minutes? (This takes place on the summer solstice on the Tropic of Cancer so that the sun is directly overhead.)	1. Determine the distance traveled by the ant's shadow: - The ant's shadow moves at a rate of 30 cm/min. - After 5 minutes, the distance traveled by the shadow is: \[ 30 \, \text{cm/min} \times 5 \, \text{min} = 150 \, \text{cm} = 1.5 \, \text{m} \] 2. Set up the problem using similar triangles: - The rope forms a right triangle with the flagpole and the ground. - The length of the rope is 10 m, the distance from the base of the flagpole to the point where the rope touches the ground is 6 m. - By the Pythagorean theorem, the height of the flagpole is: \[ \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \, \text{m} \] - The triangles formed by the ant's position on the rope and its shadow are similar to the original triangle. 3. Set up the proportion using similar triangles: - The side ratio of the similar triangles is given by the ratio of the shadow's distance to the base of the original triangle: \[ \frac{1.5 \, \text{m}}{6 \, \text{m}} = \frac{x}{8 \, \text{m}} \] - Solving for $ x $: \[ \frac{1.5}{6} = \frac{x}{8} \] \[ \frac{1.5}{6} = \frac{1}{4} = \frac{x}{8} \] \[ 4x = 8 \] \[ x = 2 \] 4. Conclusion: - The height of the ant above the ground after 5 minutes is 2 meters. The final answer is $ \boxed{2} \, \text{meters} $	2	Geometry	math-word-problem	Yes	Yes	aops_forum	false
How many rational solutions for $x$ are there to the equation $x^4+(2-p)x^3+(2-2p)x^2+(1-2p)x-p=0$ if $p$ is a prime number?	To determine the number of rational solutions for the equation \[ x^4 + (2 - p)x^3 + (2 - 2p)x^2 + (1 - 2p)x - p = 0 \] where $ p $ is a prime number, we can use the Rational Root Theorem. The Rational Root Theorem states that any rational solution, expressed as a fraction $\frac{p}{q}$, must have $ p $ as a factor of the constant term and $ q $ as a factor of the leading coefficient. 1. Identify the constant term and the leading coefficient: - The constant term is $-p$. - The leading coefficient is $1$. 2. List the possible rational roots: - The factors of the constant term $-p$ are $\pm 1, \pm p$. - The factors of the leading coefficient $1$ are $\pm 1$. - Therefore, the possible rational roots are $\pm 1, \pm p$. 3. Test the possible rational roots: - We need to check if $ x = 1, -1, p, -p $ are solutions to the polynomial equation. 4. Check $ x = 1 $: \[ 1^4 + (2 - p)1^3 + (2 - 2p)1^2 + (1 - 2p)1 - p = 1 + (2 - p) + (2 - 2p) + (1 - 2p) - p \] \[ = 1 + 2 - p + 2 - 2p + 1 - 2p - p = 4 - 6p \] Since $ 4 - 6p \neq 0 $ for any prime $ p $, $ x = 1 $ is not a solution. 5. Check $ x = -1 $: \[ (-1)^4 + (2 - p)(-1)^3 + (2 - 2p)(-1)^2 + (1 - 2p)(-1) - p = 1 - (2 - p) + (2 - 2p) - (1 - 2p) - p \] \[ = 1 - 2 + p + 2 - 2p - 1 + 2p - p = 0 \] Therefore, $ x = -1 $ is a solution. 6. Check $ x = p $: \[ p^4 + (2 - p)p^3 + (2 - 2p)p^2 + (1 - 2p)p - p = p^4 + 2p^3 - p^4 + 2p^2 - 2p^3 + p - 2p^2 - p \] \[ = p^4 - p^4 + 2p^3 - 2p^3 + 2p^2 - 2p^2 + p - p = 0 \] Therefore, $ x = p $ is a solution. 7. Check $ x = -p $: \[ (-p)^4 + (2 - p)(-p)^3 + (2 - 2p)(-p)^2 + (1 - 2p)(-p) - p = p^4 - (2 - p)p^3 + (2 - 2p)p^2 - (1 - 2p)p - p \] \[ = p^4 - 2p^3 + p^4 + 2p^2 - 2p^3 - p + 2p^2 - p \] \[ = p^4 - p^4 - 2p^3 + 2p^3 + 2p^2 - 2p^2 - p + p = 0 \] Therefore, $ x = -p $ is a solution. Since we have found that $ x = -1 $ and $ x = p $ are solutions, we conclude that there are two rational solutions. The final answer is $\boxed{2}$.	2	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] Given $x + y = 7$, find the value of x that minimizes $4x^2 + 12xy + 9y^2$. [b]p2.[/b] There are real numbers $b$ and $c$ such that the only $x$-intercept of $8y = x^2 + bx + c$ equals its $y$-intercept. Compute $b + c$. [b]p3.[/b] Consider the set of $5$ digit numbers $ABCDE$ (with $A \ne 0$) such that $A+B = C$, $B+C = D$, and $C + D = E$. What’s the size of this set? [b]p4.[/b] Let $D$ be the midpoint of $BC$ in $\vartriangle ABC$. A line perpendicular to D intersects $AB$ at $E$. If the area of $\vartriangle ABC$ is four times that of the area of $\vartriangle BDE$, what is $\angle ACB$ in degrees? [b]p5.[/b] Define the sequence $c_0, c_1, ...$ with $c_0 = 2$ and $c_k = 8c_{k-1} + 5$ for $k > 0$. Find $\lim_{k \to \infty} \frac{c_k}{8^k}$. [b]p6.[/b] Find the maximum possible value of $\|\sqrt{n^2 + 4n + 5} - \sqrt{n^2 + 2n + 5}\|$. [b]p7.[/b] Let $f(x) = \sin^8 (x) + \cos^8(x) + \frac38 \sin^4 (2x)$. Let $f^{(n)}$ (x) be the $n$th derivative of $f$. What is the largest integer $a$ such that $2^a$ divides $f^{(2020)}(15^o)$? [b]p8.[/b] Let $R^n$ be the set of vectors $(x_1, x_2, ..., x_n)$ where $x_1, x_2,..., x_n$ are all real numbers. Let $\|\|(x_1, . . . , x_n)\|\|$ denote $\sqrt{x^2_1 +... + x^2_n}$. Let $S$ be the set in $R^9$ given by $$S = \{(x, y, z) : x, y, z \in R^3 , 1 = \|\|x\|\| = \|\|y - x\|\| = \|\|z -y\|\|\}.$$ If a point $(x, y, z)$ is uniformly at random from $S$, what is $E[\|\|z\|\|^2]$? [b]p9.[/b] Let $f(x)$ be the unique integer between $0$ and $x - 1$, inclusive, that is equivalent modulo $x$ to $\left( \sum^2_{i=0} {{x-1} \choose i} ((x - 1 - i)! + i!) \right)$. Let $S$ be the set of primes between $3$ and $30$, inclusive. Find $\sum_{x\in S}^{f(x)}$. [b]p10.[/b] In the Cartesian plane, consider a box with vertices $(0, 0)$,$\left( \frac{22}{7}, 0\right)$,$(0, 24)$,$\left( \frac{22}{7}, 4\right)$. We pick an integer $a$ between $1$ and $24$, inclusive, uniformly at random. We shoot a puck from $(0, 0)$ in the direction of $\left( \frac{22}{7}, a\right)$ and the puck bounces perfectly around the box (angle in equals angle out, no friction) until it hits one of the four vertices of the box. What is the expected number of times it will hit an edge or vertex of the box, including both when it starts at $(0, 0)$ and when it ends at some vertex of the box? [b]p11.[/b] Sarah is buying school supplies and she has $\$2019$. She can only buy full packs of each of the following items. A pack of pens is $\$4$, a pack of pencils is $\$3$, and any type of notebook or stapler is $\$1$. Sarah buys at least $1$ pack of pencils. She will either buy $1$ stapler or no stapler. She will buy at most $3$ college-ruled notebooks and at most $2$ graph paper notebooks. How many ways can she buy school supplies? [b]p12.[/b] Let $O$ be the center of the circumcircle of right triangle $ABC$ with $\angle ACB = 90^o$. Let $M$ be the midpoint of minor arc $AC$ and let $N$ be a point on line $BC$ such that $MN \perp BC$. Let $P$ be the intersection of line $AN$ and the Circle $O$ and let $Q$ be the intersection of line $BP$ and $MN$. If $QN = 2$ and $BN = 8$, compute the radius of the Circle $O$. [b]p13.[/b] Reduce the following expression to a simplified rational $$\frac{1}{1 - \cos \frac{\pi}{9}}+\frac{1}{1 - \cos \frac{5 \pi}{9}}+\frac{1}{1 - \cos \frac{7 \pi}{9}}$$ [b]p14.[/b] Compute the following integral $\int_0^{\infty} \log (1 + e^{-t})dt$. [b]p15.[/b] Define $f(n)$ to be the maximum possible least-common-multiple of any sequence of positive integers which sum to $n$. Find the sum of all possible odd $f(n)$ PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To solve the problem, we need to find the number of 5-digit numbers $ABCDE$ such that the following conditions hold: 1. $A \neq 0$ 2. $A + B = C$ 3. $B + C = D$ 4. $C + D = E$ Let's start by expressing $D$ and $E$ in terms of $A$ and $B$: 1. From $A + B = C$, we have: \[ C = A + B \] 2. From $B + C = D$, substituting $C$ from the previous equation, we get: \[ D = B + (A + B) = A + 2B \] 3. From $C + D = E$, substituting $C$ and $D$ from the previous equations, we get: \[ E = (A + B) + (A + 2B) = 2A + 3B \] Now, we need to find all possible pairs $(A, B)$ such that $A$ and $B$ are digits (i.e., integers from 0 to 9) and $A \neq 0$. Since $A$ must be a non-zero digit, $A$ can take any value from 1 to 9. However, we also need to ensure that $C$, $D$, and $E$ are valid digits (i.e., between 0 and 9). Let's analyze the constraints: - $C = A + B$ must be a digit, so $0 \leq A + B \leq 9$. - $D = A + 2B$ must be a digit, so $0 \leq A + 2B \leq 9$. - $E = 2A + 3B$ must be a digit, so $0 \leq 2A + 3B \leq 9$. We will now check each possible value of $A$ and find the corresponding values of $B$ that satisfy all the constraints: 1. $A = 1$: - $1 + B \leq 9 \implies B \leq 8$ - $1 + 2B \leq 9 \implies 2B \leq 8 \implies B \leq 4$ - $2 \cdot 1 + 3B \leq 9 \implies 3B \leq 7 \implies B \leq 2$ - Possible values of $B$: $0, 1, 2$ 2. $A = 2$: - $2 + B \leq 9 \implies B \leq 7$ - $2 + 2B \leq 9 \implies 2B \leq 7 \implies B \leq 3$ - $2 \cdot 2 + 3B \leq 9 \implies 3B \leq 5 \implies B \leq 1$ - Possible values of $B$: $0, 1$ 3. $A = 3$: - $3 + B \leq 9 \implies B \leq 6$ - $3 + 2B \leq 9 \implies 2B \leq 6 \implies B \leq 3$ - $2 \cdot 3 + 3B \leq 9 \implies 3B \leq 3 \implies B \leq 1$ - Possible values of $B$: $0$ 4. $A = 4$: - $4 + B \leq 9 \implies B \leq 5$ - $4 + 2B \leq 9 \implies 2B \leq 5 \implies B \leq 2$ - $2 \cdot 4 + 3B \leq 9 \implies 3B \leq 1 \implies B \leq 0$ - Possible values of $B$: None 5. $A = 5$ to $A = 9$: - For these values of $A$, $2A + 3B$ will always exceed 9 for any non-negative $B$, so there are no valid pairs. Summarizing the valid pairs $(A, B)$: - $(1, 0), (1, 1), (1, 2)$ - $(2, 0), (2, 1)$ - $(3, 0)$ Thus, there are 6 valid pairs $(A, B)$ that satisfy all the conditions. The final answer is $\boxed{6}$	6	Algebra	math-word-problem	Yes	Yes	aops_forum	false
Let $M=\{0,1,2,\dots,2022\}$ and let $f:M\times M\to M$ such that for any $a,b\in M$, \[f(a,f(b,a))=b\] and $f(x,x)\neq x$ for each $x\in M$. How many possible functions $f$ are there $\pmod{1000}$?	1. Given the set $ M = \{0, 1, 2, \dots, 2022\} $ and the function $ f: M \times M \to M $ such that for any $ a, b \in M $, \[ f(a, f(b, a)) = b \] and $ f(x, x) \neq x $ for each $ x \in M $. 2. Denote $ c = f(a, a) $. By substituting $ a = b $ in the given functional equation, we get: \[ f(a, f(a, a)) = f(a, c) = a \] This implies that $ f(a, c) = a $. 3. Now, consider the expression $ f(c, f(a, c)) $. Since $ f(a, c) = a $, we have: \[ f(c, f(a, c)) = f(c, a) = a \] 4. Next, we use the original functional equation again with $ a $ and $ f(c, a) $: \[ f(a, f(c, a)) = f(a, a) = c \] However, we know that $ f(a, a) = c $, and by the problem's condition, $ f(x, x) \neq x $ for each $ x \in M $. This implies that $ c \neq a $. 5. We have derived that $ f(a, c) = a $ and $ f(c, a) = a $. This leads to a contradiction because if $ f(a, c) = a $ and $ f(c, a) = a $, then $ f(a, a) = c $ and $ f(c, c) = a $, which violates the condition $ f(x, x) \neq x $. 6. Therefore, no such function $ f $ can exist that satisfies both conditions simultaneously. $\blacksquare$ The final answer is $ \boxed{0} $	0	Logic and Puzzles	math-word-problem	Yes	Yes	aops_forum	false
For all positive integers $m>10^{2022}$, determine the maximum number of real solutions $x>0$ of the equation $mx=\lfloor x^{11/10}\rfloor$.	1. Setting up the problem: We are given the equation $ mx = \lfloor x^{11/10} \rfloor $ and need to determine the maximum number of real solutions $ x > 0 $ for all positive integers $ m > 10^{2022} $. 2. Initial guess and example: Consider $ m = 10^{2022} + 1 $. We can guess solutions of the form $ x = (10^{2022} + 1)^{10} + \epsilon $, where $ \epsilon $ is a small perturbation. 3. Expanding the floor function: Let $ x = k^{10} + \epsilon $ where $ k \in \mathbb{Z}^+ $ is as large as possible. We expand $ \lfloor x^{11/10} \rfloor $: \[ \lfloor (k^{10} + \epsilon)^{11/10} \rfloor = \lfloor k^{11} + \frac{11}{10} k \epsilon + \dots \rfloor = k^{11} + \lfloor \frac{11}{10} k \epsilon \rfloor \] 4. Setting up the equation: Equate this to $ mx $: \[ mx = m(k^{10} + \epsilon) \] \[ m(k^{10} + \epsilon) = k^{11} + \lfloor \frac{11}{10} k \epsilon \rfloor \] 5. Simplifying the equation: Rearrange the equation: \[ (m - k) (k^{10} + \epsilon) = \lfloor \frac{11}{10} k \epsilon \rfloor - k \epsilon \] Since $ m $ is very large, the left-hand side (LHS) is on the order of $ k^{10} $, while the right-hand side (RHS) is arbitrarily small. Therefore, both sides must equal zero: \[ \lfloor \frac{11}{10} k \epsilon \rfloor = k \epsilon \] 6. Analyzing the floor function: The equation $ \lfloor \frac{11}{10} k \epsilon \rfloor = k \epsilon $ implies that $ k \epsilon $ is an integer. Additionally, the condition $ \lfloor \frac{1}{10} k \epsilon \rfloor = 0 $ implies: \[ 0 \leq \frac{1}{10} k \epsilon < 1 \] Therefore: \[ 0 \leq k \epsilon < 10 \] 7. Counting the solutions: Since $ k \epsilon $ must be an integer and $ 0 \leq k \epsilon < 10 $, there are at most 10 possible values for $ k \epsilon $. Thus, there are at most 10 solutions for $ x = k^{10} + \epsilon $. The final answer is $\boxed{10}$	10	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
The straight line $y=ax+16$ intersects the graph of $y=x^3$ at $2$ distinct points. What is the value of $a$?	1. Set up the equations: The line $ y = ax + 16 $ intersects the curve $ y = x^3 $. To find the points of intersection, set the equations equal to each other: \[ ax + 16 = x^3 \] Rearrange this to form a polynomial equation: \[ x^3 - ax - 16 = 0 \] 2. Determine the conditions for distinct points: For the line to intersect the curve at 2 distinct points, the polynomial $ x^3 - ax - 16 = 0 $ must have exactly 2 distinct real roots. This implies that the polynomial has a double root and a single root. 3. Find the derivative: To find the double root, we need to find where the derivative of the polynomial is zero. The derivative of $ x^3 - ax - 16 $ is: \[ \frac{d}{dx}(x^3 - ax - 16) = 3x^2 - a \] Set the derivative equal to zero to find the critical points: \[ 3x^2 - a = 0 \implies a = 3x^2 \] 4. Substitute back into the original equation: Substitute $ a = 3x^2 $ into the original polynomial equation: \[ x^3 - 3x^3 - 16 = 0 \implies -2x^3 - 16 = 0 \implies x^3 = -8 \implies x = -2 \] 5. Find the value of $ a $: Substitute $ x = -2 $ back into $ a = 3x^2 $: \[ a = 3(-2)^2 = 3 \cdot 4 = 12 \] 6. Verify the solution: To ensure that the polynomial $ x^3 - 12x - 16 = 0 $ has exactly 2 distinct roots, we can check the discriminant or use the fact that the polynomial has a double root at $ x = -2 $ and another distinct root. The polynomial can be factored as: \[ (x + 2)^2(x - 4) = 0 \] This confirms that the polynomial has roots at $ x = -2 $ (double root) and $ x = 4 $ (single root). The final answer is $ \boxed{12} $	12	Algebra	math-word-problem	Yes	Yes	aops_forum	false
The bisector of $\angle BAC$ in $\triangle ABC$ intersects $BC$ in point $L$. The external bisector of $\angle ACB$ intersects $\overrightarrow{BA}$ in point $K$. If the length of $AK$ is equal to the perimeter of $\triangle ACL$, $LB=1$, and $\angle ABC=36^\circ$, find the length of $AC$.	1. Define the problem and given conditions: - We have a triangle $ \triangle ABC $. - The bisector of $ \angle BAC $ intersects $ BC $ at point $ L $. - The external bisector of $ \angle ACB $ intersects $ \overrightarrow{BA} $ at point $ K $. - The length of $ AK $ is equal to the perimeter of $ \triangle ACL $. - $ LB = 1 $. - $ \angle ABC = 36^\circ $. 2. Use the given conditions to set up the problem: - We need to find the length of $ AC $. 3. Construct auxiliary points and use the perimeter condition: - Define $ B' \in \overline{BC} $ such that $ LA = LA' $. - Define $ K' \in \overrightarrow{AC} $ such that $ AK = AK' $. - The perimeter condition implies $ CA' = CK $. 4. Analyze the angles and congruence: - Since $ \angle A'CK = \angle K'CK = 90^\circ + \frac{\gamma}{2} $, we have $ \triangle A'CK \cong \triangle K'CK $. - This congruence implies $ \angle CA'K = \angle CK'K = \frac{\alpha}{2} $. 5. Consider the position of $ A' $ relative to $ B $: - If $ A' $ is to the left of $ B $: \[ \angle C'AK < \angle CA'A = \angle LA'A = \angle LAA' < \angle LAB = \frac{\alpha}{2} \] This is a contradiction. - If $ A' $ is to the right of $ B $: \[ \angle C'AK > \angle CA'A = \angle LA'A = \angle LAA' > \angle LAB = \frac{\alpha}{2} \] This is also impossible. 6. Conclude the position of $ A' $: - Therefore, $ A' = B $ and $ \alpha = 72^\circ $. 7. Determine the length of $ AC $: - Given $ LB = 1 $ and $ \alpha = 72^\circ $, we find $ AC = AL = LB = 1 $. The final answer is $ \boxed{1} $.	1	Geometry	math-word-problem	Yes	Yes	aops_forum	false
In acute triangle $\triangle ABC$, point $R$ lies on the perpendicular bisector of $AC$ such that $\overline{CA}$ bisects $\angle BAR$. Let $Q$ be the intersection of lines $AC$ and $BR$. The circumcircle of $\triangle ARC$ intersects segment $\overline{AB}$ at $P\neq A$, with $AP=1$, $PB=5$, and $AQ=2$. Compute $AR$.	1. Identify the given information and setup the problem: - $\triangle ABC$ is an acute triangle. - Point $R$ lies on the perpendicular bisector of $AC$. - $\overline{CA}$ bisects $\angle BAR$. - $Q$ is the intersection of lines $AC$ and $BR$. - The circumcircle of $\triangle ARC$ intersects segment $\overline{AB}$ at $P \neq A$. - Given: $AP = 1$, $PB = 5$, and $AQ = 2$. 2. Analyze the cyclic quadrilateral $APCR$: - Since $APCR$ is a cyclic quadrilateral, we can use properties of cyclic quadrilaterals. - In cyclic quadrilateral $APCR$, $\angle APR = \angle ACR$ and $\angle PAR = \angle PCR$. 3. Use the given lengths to find relationships: - Given $AP = 1$ and $PB = 5$, we have $AB = AP + PB = 1 + 5 = 6$. - Given $AQ = 2$. 4. Use the fact that $R$ lies on the perpendicular bisector of $AC$: - Since $R$ is on the perpendicular bisector of $AC$, $AR = CR$. 5. Use the angle bisector theorem: - Since $\overline{CA}$ bisects $\angle BAR$, by the angle bisector theorem, $\frac{AQ}{QC} = \frac{AB}{BR}$. - Let $QC = x$. Then $\frac{2}{x} = \frac{6}{BR}$, so $BR = 3x$. 6. Use Ptolemy's theorem in cyclic quadrilateral $APCR$: - Ptolemy's theorem states that in a cyclic quadrilateral, the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals. - Applying Ptolemy's theorem to $APCR$: $AP \cdot CR + AR \cdot PC = AC \cdot PR$. - Since $AR = CR$ and $AP = 1$, we have $1 \cdot AR + AR \cdot PC = AC \cdot PR$. 7. Express $PC$ in terms of $AR$: - Since $P$ lies on $AB$, $PC = PB = 5$. 8. Solve for $AR$: - Substitute $PC = 5$ into the equation: $1 \cdot AR + AR \cdot 5 = AC \cdot PR$. - Simplify: $AR + 5AR = AC \cdot PR$. - $6AR = AC \cdot PR$. 9. Find $AC$ and $PR$: - Since $Q$ is the intersection of $AC$ and $BR$, and $AQ = 2$, $QC = x$. - $AC = AQ + QC = 2 + x$. - $PR = AR$ (since $R$ is on the perpendicular bisector of $AC$). 10. Substitute $AC$ and $PR$ into the equation: - $6AR = (2 + x) \cdot AR$. - Simplify: $6 = 2 + x$. - Solve for $x$: $x = 4$. 11. Find $AR$: - Since $AR = CR$ and $R$ is on the perpendicular bisector of $AC$, $AR = CR$. - $AR = 6$. The final answer is $\boxed{6}$.	6	Geometry	math-word-problem	Yes	Yes	aops_forum	false
Suppose that $p(x),q(x)$ are monic polynomials with nonnegative integer coefficients such that \[\frac{1}{5x}\ge\frac{1}{q(x)}-\frac{1}{p(x)}\ge\frac{1}{3x^2}\] for all integers $x\ge2$. Compute the minimum possible value of $p(1)\cdot q(1)$.	Given the inequality: \[ \frac{1}{5x} \ge \frac{1}{q(x)} - \frac{1}{p(x)} \ge \frac{1}{3x^2} \] for all integers $ x \ge 2 $, we need to find the minimum possible value of $ p(1) \cdot q(1) $. 1. Rewriting the Inequality: The inequality can be rewritten as: \[ \frac{1}{5x} \ge \frac{p(x) - q(x)}{p(x)q(x)} \ge \frac{1}{3x^2} \] Multiplying through by $ p(x)q(x) $ and rearranging, we get: \[ 3x^2(p(x) - q(x)) \ge p(x)q(x) \quad \text{(1)} \] \[ 5x(p(x) - q(x)) \le p(x)q(x) \quad \text{(2)} \] 2. Degree Analysis: From inequality (1), we see that: \[ 3x^2(p(x) - q(x)) \ge p(x)q(x) \] This implies that the degree of $ p(x) $ must be at least as large as the degree of $ q(x) $. Also, since $ p(x) $ and $ q(x) $ are monic polynomials with nonnegative integer coefficients, the degrees of $ p(x) $ and $ q(x) $ must be nonnegative integers. 3. Case Analysis: - Case 1: $ \deg q = 1 $: Let $ q(x) = x + b $. Then, $ p(x) $ must have a degree of at least 1. If $ \deg p = 2 $, then: \[ p(x) = x^2 + ax + c \] Substituting into inequality (2): \[ 5x(x^2 + ax + c - x - b) \le (x^2 + ax + c)(x + b) \] Simplifying, we get: \[ 5x(x^2 + (a-1)x + (c-b)) \le x^3 + (a+b)x^2 + (ab+c)x + bc \] Comparing the degrees, we see that this is not possible since the left-hand side has a higher degree term $ 5x^3 $ which cannot be balanced by the right-hand side. - Case 2: $ \deg q = 2 $: Let $ q(x) = x^2 + bx + c $. Then, $ p(x) $ must have a degree of at least 2. If $ \deg p = 3 $, then: \[ p(x) = x^3 + ax^2 + bx + c \] Substituting into inequality (1): \[ 3x^2(x^3 + ax^2 + bx + c - x^2 - bx - c) \ge (x^3 + ax^2 + bx + c)(x^2 + bx + c) \] Simplifying, we get: \[ 3x^2(x^3 + (a-1)x^2 + (b-b)x + (c-c)) \ge x^5 + (a+b)x^4 + (ab+c)x^3 + bcx^2 \] Comparing the degrees, we see that this is not possible since the left-hand side has a higher degree term $ 3x^5 $ which cannot be balanced by the right-hand side. 4. Finding the Minimum Value: After eliminating the impossible cases, we need to find the polynomials $ p(x) $ and $ q(x) $ that satisfy the given inequality and minimize $ p(1) \cdot q(1) $. - Let $ q(x) = x^2 + 2x $ and $ p(x) = x^5 $. Then: \[ p(1) = 1^5 = 1 \] \[ q(1) = 1^2 + 2 \cdot 1 = 3 \] Therefore: \[ p(1) \cdot q(1) = 1 \cdot 3 = 3 \] The final answer is $ \boxed{3} $.	3	Inequalities	math-word-problem	Yes	Yes	aops_forum	false
Michelle is drawing segments in the plane. She begins from the origin facing up the $y$-axis and draws a segment of length $1$. Now, she rotates her direction by $120^\circ$, with equal probability clockwise or counterclockwise, and draws another segment of length $1$ beginning from the end of the previous segment. She then continues this until she hits an already drawn segment. What is the expected number of segments she has drawn when this happens?	1. Understanding the Problem: Michelle starts at the origin and draws segments of length 1. After each segment, she rotates by $120^\circ$ either clockwise or counterclockwise with equal probability. We need to find the expected number of segments she draws before hitting an already drawn segment. 2. Initial Segments: The first segment is drawn from the origin along the positive $y$-axis. The second segment is drawn after rotating $120^\circ$ either clockwise or counterclockwise. These two segments do not intersect. 3. Probability of Hitting a Segment: From the third segment onward, there is a probability of $\frac{1}{2}$ that the new segment will intersect an already drawn segment. This is because each new segment can either intersect or not intersect the previous segments with equal probability. 4. Expected Number of Segments: Let $S$ be the expected number of segments drawn before hitting an already drawn segment. We can express $S$ as a series: \[ S = \sum_{n=3}^{\infty} n \cdot \left(\frac{1}{2}\right)^{n-2} \] This series accounts for the fact that starting from the third segment, each segment has a $\frac{1}{2}$ probability of hitting an already drawn segment. 5. Simplifying the Series: To simplify the series, we can use the method of subtracting a shifted version of the series from itself. First, write the series: \[ S = \frac{3}{2} + \frac{4}{2^2} + \frac{5}{2^3} + \frac{6}{2^4} + \cdots \] Now, consider halving the series: \[ \frac{S}{2} = \frac{3}{2^2} + \frac{4}{2^3} + \frac{5}{2^4} + \cdots \] Subtracting $\frac{S}{2}$ from $S$: \[ S - \frac{S}{2} = \left(\frac{3}{2} + \frac{4}{2^2} + \frac{5}{2^3} + \cdots\right) - \left(\frac{3}{2^2} + \frac{4}{2^3} + \frac{5}{2^4} + \cdots\right) \] This simplifies to: \[ \frac{S}{2} = \frac{3}{2} + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots \] The series on the right is a geometric series with the first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$: \[ \frac{S}{2} = \frac{3}{2} + \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots\right) \] The sum of the geometric series is: \[ \sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^{k+1} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1 \] Therefore: \[ \frac{S}{2} = \frac{3}{2} + 1 = \frac{5}{2} \] Solving for $S$: \[ S = 2 \cdot \frac{5}{2} = 5 \] The final answer is $\boxed{5}$.	5	Geometry	math-word-problem	Yes	Yes	aops_forum	false
We say that an integer $x\in\{1,\dots,102\}$ is $\textit{square-ish}$ if there exists some integer $n$ such that $x\equiv n^2+n\pmod{103}$. Compute the product of all $\textit{square-ish}$ integers modulo $103$.	1. Understanding the Problem: We need to find the product of all integers $ x \in \{1, \dots, 102\} $ that can be expressed as $ x \equiv n^2 + n \pmod{103} $ for some integer $ n $. 2. Analyzing the Condition: The condition $ x \equiv n^2 + n \pmod{103} $ can be rewritten as: \[ x \equiv n(n + 1) \pmod{103} \] We need to determine the product of all such $ x $. 3. Identifying Distinct Values: For distinct $ m $ and $ n $, the values $ m^2 + m $ and $ n^2 + n $ are distinct modulo 103 if $ m + n \not\equiv -1 \pmod{103} $. This means that each $ n $ from 0 to 102 will give a unique value of $ n(n + 1) $ modulo 103. 4. Range of $ n $: Since $ n $ ranges from 0 to 102, we need to consider the product of $ n(n + 1) $ for $ n = 0, 1, 2, \ldots, 102 $. 5. Simplifying the Product: The product of all $ n(n + 1) $ for $ n = 0 $ to $ 102 $ can be written as: \[ \prod_{n=0}^{102} n(n + 1) \] This can be split into two separate products: \[ \left( \prod_{n=0}^{102} n \right) \left( \prod_{n=0}^{102} (n + 1) \right) \] 6. Calculating the Factorials: The first product is the factorial of 102: \[ \prod_{n=0}^{102} n = 102! \] The second product is the factorial of 103: \[ \prod_{n=0}^{102} (n + 1) = 103! \] 7. Using Wilson's Theorem: Wilson's Theorem states that for a prime $ p $: \[ (p-1)! \equiv -1 \pmod{p} \] Applying this to 103, we get: \[ 102! \equiv -1 \pmod{103} \] And since $ 103! = 103 \cdot 102! $: \[ 103! \equiv 0 \cdot (-1) \equiv 0 \pmod{103} \] 8. Combining the Results: The product of all $ n(n + 1) $ modulo 103 is: \[ 102! \cdot 103! \equiv (-1) \cdot 0 \equiv 0 \pmod{103} \] The final answer is $\boxed{0}$.	0	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Suppose $a$ and $b$ are positive integers with a curious property: $(a^3 - 3ab +\tfrac{1}{2})^n + (b^3 +\tfrac{1}{2})^n$ is an integer for at least $3$, but at most finitely many different choices of positive integers $n$. What is the least possible value of $a+b$?	1. We start with the given expression $(a^3 - 3ab + \tfrac{1}{2})^n + (b^3 + \tfrac{1}{2})^n$ and note that it must be an integer for at least 3, but at most finitely many different choices of positive integers $n$. 2. To simplify the problem, we multiply the entire expression by 2 to clear the fractions: \[ (2a^3 - 6ab + 1)^n + (2b^3 + 1)^n \] This expression must be an integer for the same values of $n$. 3. For the expression to be an integer, it must be divisible by $2^n$. We analyze the expression modulo 4: \[ (2a^3 - 6ab + 1)^n + (2b^3 + 1)^n \pmod{4} \] Since $2a^3 - 6ab + 1$ and $2b^3 + 1$ are both odd, their sum modulo 4 is: \[ (2a^3 - 6ab + 1) + (2b^3 + 1) = 2(a^3 + b^3 - 3ab + 1) \] This must be divisible by $2^n$ for at least 3 values of $n$. 4. We need $2(a^3 + b^3 - 3ab + 1)$ to be divisible by $2^n$ for $n \geq 3$. This implies: \[ a^3 + b^3 - 3ab + 1 \equiv 0 \pmod{8} \] 5. We now factorize $a^3 + b^3 - 3ab + 1$: \[ a^3 + b^3 - 3ab + 1 = (a + b + 1)(a^2 + b^2 - ab - a - b + 1) \] We need this product to be divisible by 8. 6. We consider the parity of $a$ and $b$. Since $a$ and $b$ are positive integers, $a + b + 1$ and $a^2 + b^2 - ab - a - b + 1$ must have opposite parity for their product to be divisible by 8. 7. We test small values of $a$ and $b$ to find the smallest $a + b$ such that $a^3 + b^3 - 3ab + 1$ is divisible by 8: - For $a = 1$ and $b = 5$: \[ a^3 + b^3 - 3ab + 1 = 1^3 + 5^3 - 3 \cdot 1 \cdot 5 + 1 = 1 + 125 - 15 + 1 = 112 \] \[ 112 \div 8 = 14 \quad \text{(divisible by 8)} \] - Therefore, $a = 1$ and $b = 5$ is a valid pair. 8. The least possible value of $a + b$ is: \[ a + b = 1 + 5 = 6 \] The final answer is $\boxed{6}$.	6	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
[b]p7.[/b] An ant starts at the point $(0, 0)$. It travels along the integer lattice, at each lattice point choosing the positive $x$ or $y$ direction with equal probability. If the ant reaches $(20, 23)$, what is the probability it did not pass through $(20, 20)$? [b]p8.[/b] Let $a_0 = 2023$ and $a_n$ be the sum of all divisors of $a_{n-1}$ for all $n \ge 1$. Compute the sum of the prime numbers that divide $a_3$. [b]p9.[/b] Five circles of radius one are stored in a box of base length five as in the following diagram. How far above the base of the box are the upper circles touching the sides of the box? [img]https://cdn.artofproblemsolving.com/attachments/7/c/c20b5fa21fbd8ce791358fd888ed78fcdb7646.png[/img] PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To solve the problem, we need to compute the sum of the prime numbers that divide $a_3$. We start with $a_0 = 2023$ and use the given recurrence relation $a_n$ which is the sum of all divisors of $a_{n-1}$. 1. Compute $a_1$: \[ a_0 = 2023 = 7 \cdot 17^2 \] The sum of the divisors of $2023$ is: \[ a_1 = (1 + 7)(1 + 17 + 17^2) = 8 \cdot (1 + 17 + 289) = 8 \cdot 307 = 2456 \] 2. Compute $a_2$: \[ 2456 = 2^3 \cdot 307 \] The sum of the divisors of $2456$ is: \[ a_2 = (1 + 2 + 4 + 8)(1 + 307) = 15 \cdot 308 = 4620 \] 3. Compute $a_3$: \[ 4620 = 2^2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \] The sum of the divisors of $4620$ is: \[ a_3 = (1 + 2 + 4)(1 + 3)(1 + 5)(1 + 7)(1 + 11) \] Calculate each term: \[ 1 + 2 + 4 = 7 \] \[ 1 + 3 = 4 \] \[ 1 + 5 = 6 \] \[ 1 + 7 = 8 \] \[ 1 + 11 = 12 \] Therefore: \[ a_3 = 7 \cdot 4 \cdot 6 \cdot 8 \cdot 12 \] Simplify the product: \[ a_3 = 7 \cdot (2^2) \cdot (2 \cdot 3) \cdot (2^3) \cdot (2^2 \cdot 3) = 2^8 \cdot 3^2 \cdot 7 \] 4. Sum of the prime numbers that divide $a_3$: The prime factors of $a_3$ are $2$, $3$, and $7$. Their sum is: \[ 2 + 3 + 7 = 12 \] The final answer is $\boxed{12}$	12	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
[b]p10.[/b] Three rectangles of dimension $X \times 2$ and four rectangles of dimension $Y \times 1$ are the pieces that form a rectangle of area $3XY$ where $X$ and $Y$ are positive, integer values. What is the sum of all possible values of $X$? [b]p11.[/b] Suppose we have a polynomial $p(x) = x^2 + ax + b$ with real coefficients $a + b = 1000$ and $b > 0$. Find the smallest possible value of $b$ such that $p(x)$ has two integer roots. [b]p12.[/b] Ten square slips of paper of the same size, numbered $0, 1, 2, ..., 9$, are placed into a bag. Four of these squares are then randomly chosen and placed into a two-by-two grid of squares. What is the probability that the numbers in every pair of blocks sharing a side have an absolute difference no greater than two? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. We start with the given equation for the area of the rectangle formed by the pieces: \[ 3 \cdot (X \times 2) + 4 \cdot (Y \times 1) = 3XY \] Simplifying, we get: \[ 6X + 4Y = 3XY \] 2. Rearrange the equation to isolate terms involving $X$ and $Y$: \[ 3XY - 6X - 4Y = 0 \] Add 8 to both sides to facilitate factoring: \[ 3XY - 6X - 4Y + 8 = 8 \] 3. Factor the left-hand side: \[ (3X - 4)(Y - 2) = 8 \] 4. Since $X$ and $Y$ are positive integers, we need to find pairs $(3X - 4, Y - 2)$ that multiply to 8. The factor pairs of 8 are: \[ (1, 8), (2, 4), (4, 2), (8, 1) \] 5. Solve for $X$ and $Y$ for each pair: - For $(3X - 4, Y - 2) = (1, 8)$: \[ 3X - 4 = 1 \implies 3X = 5 \implies X = \frac{5}{3} \quad (\text{not an integer}) \] - For $(3X - 4, Y - 2) = (2, 4)$: \[ 3X - 4 = 2 \implies 3X = 6 \implies X = 2 \] \[ Y - 2 = 4 \implies Y = 6 \] - For $(3X - 4, Y - 2) = (4, 2)$: \[ 3X - 4 = 4 \implies 3X = 8 \implies X = \frac{8}{3} \quad (\text{not an integer}) \] - For $(3X - 4, Y - 2) = (8, 1)$: \[ 3X - 4 = 8 \implies 3X = 12 \implies X = 4 \] \[ Y - 2 = 1 \implies Y = 3 \] 6. The possible integer values of $X$ are 2 and 4. Summing these values: \[ 2 + 4 = 6 \] The final answer is $\boxed{6}$.	6	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
[b]p19.[/b] $A_1A_2...A_{12}$ is a regular dodecagon with side length $1$ and center at point $O$. What is the area of the region covered by circles $(A_1A_2O)$, $(A_3A_4O)$, $(A_5A_6O)$, $(A_7A_8O)$, $(A_9A_{10}O)$, and $(A_{11}A_{12}O)$? $(ABC)$ denotes the circle passing through points $A,B$, and $C$. [b]p20.[/b] Let $N = 2000... 0x0 ... 00023$ be a $2023$-digit number where the $x$ is the $23$rd digit from the right. If$ N$ is divisible by $13$, compute $x$. [b]p21.[/b] Alice and Bob each visit the dining hall to get a grilled cheese at a uniformly random time between $12$ PM and $1$ PM (their arrival times are independent) and, after arrival, will wait there for a uniformly random amount of time between $0$ and $30$ minutes. What is the probability that they will meet? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. Let $ N' = 200\ldots0023 $, where there are $ 2020 $ zeroes. We need to find $ x $ such that $ N = N' + x \cdot 10^{2000} $ is divisible by $ 13 $. 2. First, we compute $ N' \mod 13 $: \[ N' = 2 \cdot 10^{2022} + 23 \] Since $ 10^3 \equiv -1 \pmod{13} $, we have: \[ 10^{2022} = (10^3)^{674} \equiv (-1)^{674} \equiv 1 \pmod{13} \] Therefore: \[ 2 \cdot 10^{2022} \equiv 2 \cdot 1 \equiv 2 \pmod{13} \] Adding the constant term: \[ N' \equiv 2 + 23 \equiv 2 + 10 \equiv 12 \pmod{13} \] 3. Next, we compute $ 10^{2000} \mod 13 $: \[ 10^{2000} = (10^3)^{666} \cdot 10^2 \equiv (-1)^{666} \cdot 100 \equiv 1 \cdot 100 \equiv 100 \pmod{13} \] Since $ 100 \equiv 9 \pmod{13} $: \[ 10^{2000} \equiv 9 \pmod{13} \] 4. Now, we combine the results: \[ N \equiv N' + x \cdot 10^{2000} \pmod{13} \] Substituting the values: \[ 0 \equiv 12 + 9x \pmod{13} \] Solving for $ x $: \[ 9x \equiv -12 \pmod{13} \] Since $ -12 \equiv 1 \pmod{13} $: \[ 9x \equiv 1 \pmod{13} \] To find $ x $, we need the multiplicative inverse of $ 9 $ modulo $ 13 $. The inverse of $ 9 $ modulo $ 13 $ is $ 3 $ because: \[ 9 \cdot 3 \equiv 27 \equiv 1 \pmod{13} \] Therefore: \[ x \equiv 3 \pmod{13} \] The final answer is $ \boxed{3} $	3	Geometry	math-word-problem	Yes	Yes	aops_forum	false
A sequence $a_1, a_2, \ldots$ satisfies $a_1 = \dfrac 52$ and $a_{n + 1} = {a_n}^2 - 2$ for all $n \ge 1.$ Let $M$ be the integer which is closest to $a_{2023}.$ The last digit of $M$ equals $$ \mathrm a. ~ 0\qquad \mathrm b.~2\qquad \mathrm c. ~4 \qquad \mathrm d. ~6 \qquad \mathrm e. ~8 $$	1. We start with the given sequence $a_1, a_2, \ldots$ defined by $a_1 = \frac{5}{2}$ and $a_{n+1} = a_n^2 - 2$ for all $n \geq 1$. 2. We observe that the sequence can be transformed using the substitution $a_n = u_n + \frac{1}{u_n}$. This substitution simplifies the recurrence relation. Let's verify this: \[ a_{n+1} = a_n^2 - 2 = \left(u_n + \frac{1}{u_n}\right)^2 - 2 \] \[ a_{n+1} = u_n^2 + 2 + \frac{1}{u_n^2} - 2 = u_n^2 + \frac{1}{u_n^2} \] This shows that if $a_n = u_n + \frac{1}{u_n}$, then $a_{n+1} = u_n^2 + \frac{1}{u_n^2}$. 3. Given $a_1 = \frac{5}{2}$, we can find $u_1$ such that: \[ u_1 + \frac{1}{u_1} = \frac{5}{2} \] Solving this quadratic equation: \[ 2u_1^2 - 5u_1 + 2 = 0 \] Using the quadratic formula $u_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: \[ u_1 = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \] \[ u_1 = 2 \quad \text{or} \quad u_1 = \frac{1}{2} \] We choose $u_1 = 2$ (since it simplifies the calculations). 4. Now, we have $u_1 = 2$ and the recurrence relation $u_{n+1} = u_n^2$. Therefore: \[ u_2 = u_1^2 = 2^2 = 4 \] \[ u_3 = u_2^2 = 4^2 = 16 \] \[ u_4 = u_3^2 = 16^2 = 256 \] Continuing this pattern, we see that $u_n = 2^{2^{n-1}}$. 5. For $a_{2023}$, we have: \[ u_{2023} = 2^{2^{2022}} \] \[ a_{2023} = u_{2023} + \frac{1}{u_{2023}} = 2^{2^{2022}} + 2^{-2^{2022}} \] Since $2^{-2^{2022}}$ is extremely small, $a_{2023}$ is very close to $2^{2^{2022}}$. 6. Let $M$ be the integer closest to $a_{2023}$. Thus, $M \approx 2^{2^{2022}}$. 7. To find the last digit of $M$, we need to determine the last digit of $2^{2^{2022}}$. The last digit of powers of 2 cycles every 4: \[ 2^1 \equiv 2 \pmod{10} \] \[ 2^2 \equiv 4 \pmod{10} \] \[ 2^3 \equiv 8 \pmod{10} \] \[ 2^4 \equiv 6 \pmod{10} \] \[ 2^5 \equiv 2 \pmod{10} \] and so on. 8. Since $2^{2022} \equiv 2 \pmod{4}$ (because $2022 \mod 4 = 2$), we have: \[ 2^{2^{2022}} \equiv 2^2 \equiv 4 \pmod{10} \] The final answer is $\boxed{4}$	4	Other	MCQ	Yes	Yes	aops_forum	false
How many ordered triples of integers $(a, b, c)$ satisfy the following system? $$ \begin{cases} ab + c &= 17 \\ a + bc &= 19 \end{cases} $$ $$ \mathrm a. ~ 2\qquad \mathrm b.~3\qquad \mathrm c. ~4 \qquad \mathrm d. ~5 \qquad \mathrm e. ~6 $$	To solve the system of equations: \[ \begin{cases} ab + c &= 17 \\ a + bc &= 19 \end{cases} \] 1. Subtract the two equations: \[ (ab + c) - (a + bc) = 17 - 19 \] Simplifying, we get: \[ ab + c - a - bc = -2 \] Rearrange terms: \[ a(b-1) - c(b-1) = -2 \] Factor out $(b-1)$: \[ (a - c)(b - 1) = -2 \] 2. Add the two equations: \[ (ab + c) + (a + bc) = 17 + 19 \] Simplifying, we get: \[ ab + c + a + bc = 36 \] Rearrange terms: \[ a(1 + b) + c(1 + b) = 36 \] Factor out $(1 + b)$: \[ (a + c)(b + 1) = 36 \] 3. Analyze the factorizations: We now have two key equations: \[ (a - c)(b - 1) = -2 \] \[ (a + c)(b + 1) = 36 \] 4. Find integer solutions: We need to find integer values of $a$, $b$, and $c$ that satisfy both equations. We will consider the possible factorizations of $-2$ and $36$. - For $(a - c)(b - 1) = -2$, the possible pairs $(a - c, b - 1)$ are: \[ (-1, 2), (1, -2), (-2, 1), (2, -1) \] - For $(a + c)(b + 1) = 36$, the possible pairs $(a + c, b + 1)$ are: \[ (1, 36), (2, 18), (3, 12), (4, 9), (6, 6), (-1, -36), (-2, -18), (-3, -12), (-4, -9), (-6, -6) \] 5. Match pairs: We need to find pairs $(a, c)$ and $b$ that satisfy both equations. We will test each pair: - For $(a - c, b - 1) = (-1, 2)$: \[ b = 3 \quad \text{and} \quad (a + c)(3 + 1) = 36 \implies (a + c)4 = 36 \implies a + c = 9 \] \[ a - c = -1 \] Solving these: \[ a + c = 9 \quad \text{and} \quad a - c = -1 \] Adding these: \[ 2a = 8 \implies a = 4 \] Subtracting these: \[ 2c = 10 \implies c = 5 \] So, $(a, b, c) = (4, 3, 5)$. - For $(a - c, b - 1) = (1, -2)$: \[ b = -1 \quad \text{and} \quad (a + c)(-1 + 1) = 36 \implies (a + c)0 = 36 \implies \text{no solution} \] - For $(a - c, b - 1) = (-2, 1)$: \[ b = 2 \quad \text{and} \quad (a + c)(2 + 1) = 36 \implies (a + c)3 = 36 \implies a + c = 12 \] \[ a - c = -2 \] Solving these: \[ a + c = 12 \quad \text{and} \quad a - c = -2 \] Adding these: \[ 2a = 10 \implies a = 5 \] Subtracting these: \[ 2c = 14 \implies c = 7 \] So, $(a, b, c) = (5, 2, 7)$. - For $(a - c, b - 1) = (2, -1)$: \[ b = 0 \quad \text{and} \quad (a + c)(0 + 1) = 36 \implies a + c = 36 \] \[ a - c = 2 \] Solving these: \[ a + c = 36 \quad \text{and} \quad a - c = 2 \] Adding these: \[ 2a = 38 \implies a = 19 \] Subtracting these: \[ 2c = 34 \implies c = 17 \] So, $(a, b, c) = (19, 0, 17)$. 6. Conclusion: The valid ordered triples $(a, b, c)$ are: \[ (4, 3, 5), (5, 2, 7), (19, 0, 17) \] The final answer is $\boxed{3}$	3	Algebra	MCQ	Yes	Yes	aops_forum	false
What is the least positive integer $m$ such that the following is true? [i]Given $\it m$ integers between $\it1$ and $\it{2023},$ inclusive, there must exist two of them $\it a, b$ such that $1 < \frac ab \le 2.$ [/i] \[\mathrm a. ~ 10\qquad \mathrm b.~11\qquad \mathrm c. ~12 \qquad \mathrm d. ~13 \qquad \mathrm e. ~1415\]	To solve this problem, we need to find the smallest positive integer $ m $ such that among any $ m $ integers chosen from the set $\{1, 2, \ldots, 2023\}$, there exist two integers $ a $ and $ b $ satisfying $ 1 < \frac{a}{b} \leq 2 $. 1. Understanding the Problem: - We need to ensure that for any $ m $ integers chosen from $\{1, 2, \ldots, 2023\}$, there are two integers $ a $ and $ b $ such that $ 1 < \frac{a}{b} \leq 2 $. - This means $ a $ must be at most twice $ b $ but greater than $ b $. 2. Constructing a Sequence: - To avoid having $ 1 < \frac{a}{b} \leq 2 $, we need to construct a sequence where each number is more than twice the previous number. - Start with the smallest number, say $ a_1 = 1 $. The next number $ a_2 $ must be greater than $ 2 \times 1 = 2 $, so $ a_2 \geq 3 $. - Continue this process: $ a_3 \geq 2 \times 3 = 6 $, $ a_4 \geq 2 \times 6 = 12 $, and so on. 3. Finding the Maximum Sequence: - We need to find the maximum number of terms in this sequence that can fit within the range $\{1, 2, \ldots, 2023\}$. - Start with $ a_1 = 1 $: \[ \begin{align} a_1 &= 1, \\ a_2 &= 3, \\ a_3 &= 6, \\ a_4 &= 12, \\ a_5 &= 24, \\ a_6 &= 48, \\ a_7 &= 96, \\ a_8 &= 192, \\ a_9 &= 384, \\ a_{10} &= 768, \\ a_{11} &= 1536. \end{align} \] - The next term would be $ a_{12} = 2 \times 1536 = 3072 $, which is outside the range $\{1, 2, \ldots, 2023\}$. 4. Conclusion: - We can have at most 11 terms in this sequence without violating the condition $ 1 < \frac{a}{b} \leq 2 $. - Therefore, if we choose 12 numbers, by the pigeonhole principle, there must be at least two numbers $ a $ and $ b $ such that $ 1 < \frac{a}{b} \leq 2 $. The final answer is $\boxed{12}$.	12	Number Theory	MCQ	Yes	Yes	aops_forum	false
[b]p1.[/b] Twelve tables are set up in a row for a Millenium party. You want to put $2000$ cupcakes on the tables so that the numbers of cupcakes on adjacent tables always differ by one (for example, if the $5$th table has $20$ cupcakes, then the $4$th table has either $19$ or $21$ cupcakes, and the $6$th table has either $19$ or $21$ cupcakes). a) Find a way to do this. b) Suppose a Y2K bug eats one of the cupcakes, so you have only $1999$ cupcakes. Show that it is impossible to arrange the cupcakes on the tables according to the above conditions. [b]p2.[/b] Let $P$ and $Q$ lie on the hypotenuse $AB$ of the right triangle $CAB$ so that $\|AP\|=\|PQ\|=\|QB\|=\|AB\|/3$. Suppose that $\|CP\|^2+\|CQ\|^2=5$. Prove that $\|AB\|$ has the same value for all such triangles, and find that value. Note: $\|XY\|$ denotes the length of the segment $XY$. [b]p3.[/b] Let $P$ be a polynomial with integer coefficients and let $a, b, c$ be integers. Suppose $P(a)=b$, $P(b)=c$, and $P(c)=a$. Prove that $a=b=c$. [b]p4.[/b] A lattice point is a point $(x,y)$ in the plane for which both $x$ and $y$ are integers. Each lattice point is painted with one of $1999$ available colors. Prove that there is a rectangle (of nonzero height and width) whose corners are lattice points of the same color. [b]p5.[/b] A $1999$-by-$1999$ chocolate bar has vertical and horizontal grooves which divide it into $1999^2$ one-by-one squares. Caesar and Brutus are playing the following game with the chocolate bar: A move consists of a player picking up one chocolate rectangle; breaking it along a groove into two smaller rectangles; and then either putting both rectangles down or eating one piece and putting the other piece down. The players move alternately. The one who cannot make a move at his turn (because there are only one-by-one squares left) loses. Caesar starts. Which player has a winning strategy? Describe a winning strategy for that player. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To solve this problem, we will use coordinate geometry to find the value of $ \|AB\| $. 1. Assign coordinates to points: Let $ C = (0, 0) $, $ B = (3b, 0) $, and $ A = (0, 3a) $. This setup places $ C $ at the origin, $ B $ on the x-axis, and $ A $ on the y-axis. 2. Determine coordinates of $ P $ and $ Q $: Since $ \|AP\| = \|PQ\| = \|QB\| = \frac{\|AB\|}{3} $, we can place $ P $ and $ Q $ as follows: - $ P $ is one-third of the way from $ A $ to $ B $, so $ P = (b, 2a) $. - $ Q $ is two-thirds of the way from $ A $ to $ B $, so $ Q = (2b, a) $. 3. Calculate distances $ \|CP\| $ and $ \|CQ\| $: Using the distance formula, we find: \[ \|CP\| = \sqrt{(b - 0)^2 + (2a - 0)^2} = \sqrt{b^2 + 4a^2} \] \[ \|CQ\| = \sqrt{(2b - 0)^2 + (a - 0)^2} = \sqrt{4b^2 + a^2} \] 4. Use the given condition $ \|CP\|^2 + \|CQ\|^2 = 5 $: Substitute the distances into the given condition: \[ (\|CP\|)^2 + (\|CQ\|)^2 = (b^2 + 4a^2) + (4b^2 + a^2) = 5 \] Simplify the equation: \[ b^2 + 4a^2 + 4b^2 + a^2 = 5 \] \[ 5b^2 + 5a^2 = 5 \] \[ b^2 + a^2 = 1 \] 5. Calculate $ \|AB\| $: The length of $ AB $ is given by: \[ \|AB\| = \sqrt{(3b - 0)^2 + (0 - 3a)^2} = \sqrt{9b^2 + 9a^2} \] Substitute $ b^2 + a^2 = 1 $: \[ \|AB\| = \sqrt{9(b^2 + a^2)} = \sqrt{9 \cdot 1} = \sqrt{9} = 3 \] Thus, the value of $ \|AB\| $ is always 3 for any right triangle $ CAB $ with the given conditions. The final answer is $ \boxed{3} $	3	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
In a convex pentagon $ABCDE$ the sides have lengths $1,2,3,4,$ and $5$, though not necessarily in that order. Let $F,G,H,$ and $I$ be the midpoints of the sides $AB$, $BC$, $CD$, and $DE$, respectively. Let $X$ be the midpoint of segment $FH$, and $Y$ be the midpoint of segment $GI$. The length of segment $XY$ is an integer. Find all possible values for the length of side $AE$.	1. Assign coordinates to the vertices of the pentagon: Let $ A \equiv (0,0) $, $ B \equiv (4\alpha_1, 4\beta_1) $, $ C \equiv (4\alpha_2, 4\beta_2) $, $ D \equiv (4\alpha_3, 4\beta_3) $, and $ E \equiv (4\alpha_4, 4\beta_4) $. 2. Find the coordinates of the midpoints: - $ F $ is the midpoint of $ AB $: \[ F = \left( \frac{0 + 4\alpha_1}{2}, \frac{0 + 4\beta_1}{2} \right) = (2\alpha_1, 2\beta_1) \] - $ G $ is the midpoint of $ BC $: \[ G = \left( \frac{4\alpha_1 + 4\alpha_2}{2}, \frac{4\beta_1 + 4\beta_2}{2} \right) = (2(\alpha_1 + \alpha_2), 2(\beta_1 + \beta_2)) \] - $ H $ is the midpoint of $ CD $: \[ H = \left( \frac{4\alpha_2 + 4\alpha_3}{2}, \frac{4\beta_2 + 4\beta_3}{2} \right) = (2(\alpha_2 + \alpha_3), 2(\beta_2 + \beta_3)) \] - $ I $ is the midpoint of $ DE $: \[ I = \left( \frac{4\alpha_3 + 4\alpha_4}{2}, \frac{4\beta_3 + 4\beta_4}{2} \right) = (2(\alpha_3 + \alpha_4), 2(\beta_3 + \beta_4)) \] 3. Find the coordinates of $ X $ and $ Y $: - $ X $ is the midpoint of $ FH $: \[ X = \left( \frac{2\alpha_1 + 2(\alpha_2 + \alpha_3)}{2}, \frac{2\beta_1 + 2(\beta_2 + \beta_3)}{2} \right) = (\alpha_1 + \alpha_2 + \alpha_3, \beta_1 + \beta_2 + \beta_3) \] - $ Y $ is the midpoint of $ GI $: \[ Y = \left( \frac{2(\alpha_1 + \alpha_2) + 2(\alpha_3 + \alpha_4)}{2}, \frac{2(\beta_1 + \beta_2) + 2(\beta_3 + \beta_4)}{2} \right) = (\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4, \beta_1 + \beta_2 + \beta_3 + \beta_4) \] 4. Calculate the length of segment $ XY $: \[ XY = \sqrt{(\alpha_4)^2 + (\beta_4)^2} \] Since $ XY $ is given to be an integer, $ \sqrt{(\alpha_4)^2 + (\beta_4)^2} $ must be an integer. 5. Relate $ XY $ to $ AE $: The length of side $ AE $ is: \[ AE = 4 \sqrt{(\alpha_4)^2 + (\beta_4)^2} \] Since $ \sqrt{(\alpha_4)^2 + (\beta_4)^2} $ is an integer, let $ k = \sqrt{(\alpha_4)^2 + (\beta_4)^2} $. Therefore, $ AE = 4k $. 6. Determine possible values for $ AE $: Given the side lengths of the pentagon are $ 1, 2, 3, 4, $ and $ 5 $, and $ AE = 4k $, the only possible value for $ AE $ that fits within these lengths is $ 4 $. Conclusion: \[ \boxed{4} \]	4	Geometry	math-word-problem	Yes	Yes	aops_forum	false
Let $r$ be a nonzero real number. The values of $z$ which satisfy the equation \[ r^4z^4 + (10r^6-2r^2)z^2-16r^5z+(9r^8+10r^4+1) = 0 \] are plotted on the complex plane (i.e. using the real part of each root as the x-coordinate and the imaginary part as the y-coordinate). Show that the area of the convex quadrilateral with these points as vertices is independent of $r$, and find this area.	1. We start with the given quartic equation: \[ r^4z^4 + (10r^6 - 2r^2)z^2 - 16r^5z + (9r^8 + 10r^4 + 1) = 0 \] 2. We factor the quartic polynomial. Notice that the constant term can be factored as: \[ 9r^8 + 10r^4 + 1 = (9r^4 + 1)(r^4 + 1) \] 3. Assume the quartic can be factored into two quadratics: \[ (r^2z^2 + az + b)(r^2z^2 + cz + d) \] 4. By comparing coefficients, we find: \[ r^4z^4 + (a+c)r^2z^3 + (ac + b + d)z^2 + (ad + bc)z + bd = r^4z^4 + (10r^6 - 2r^2)z^2 - 16r^5z + (9r^8 + 10r^4 + 1) \] 5. From the constant term, we have: \[ bd = 9r^8 + 10r^4 + 1 \] 6. From the coefficient of $z$, we have: \[ ad + bc = -16r^5 \] 7. From the coefficient of $z^2$, we have: \[ ac + b + d = 10r^6 - 2r^2 \] 8. We can factor the quartic as: \[ (r^2z^2 + 2rz + (9r^4 + 1))(r^2z^2 - 2rz + (r^4 + 1)) \] 9. Using the quadratic formula on each quadratic factor, we find the roots: \[ r^2z^2 + 2rz + (9r^4 + 1) = 0 \implies z = \frac{-2r \pm \sqrt{4r^2 - 4r^2(9r^4 + 1)}}{2r^2} = \frac{-2r \pm 2r\sqrt{9r^4}}{2r^2} = \frac{-1 \pm 3ri}{r} \] \[ r^2z^2 - 2rz + (r^4 + 1) = 0 \implies z = \frac{2r \pm \sqrt{4r^2 - 4r^2(r^4 + 1)}}{2r^2} = \frac{2r \pm 2r\sqrt{r^4}}{2r^2} = \frac{1 \pm ri}{r} \] 10. The roots are: \[ z_1 = \frac{-1 + 3ri}{r}, \quad z_2 = \frac{-1 - 3ri}{r}, \quad z_3 = \frac{1 + ri}{r}, \quad z_4 = \frac{1 - ri}{r} \] 11. Plotting these points on the complex plane, we have: \[ z_1 = -\frac{1}{r} + 3i, \quad z_2 = -\frac{1}{r} - 3i, \quad z_3 = \frac{1}{r} + i, \quad z_4 = \frac{1}{r} - i \] 12. These points form an isosceles trapezoid with height $\frac{2}{r}$ and bases $2ri$ and $6ri$. 13. The area of the trapezoid is given by: \[ \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} = \frac{1}{2} \times (2r + 6r) \times \frac{2}{r} = \frac{1}{2} \times 8r \times \frac{2}{r} = 8 \] The final answer is $\boxed{8}$.	8	Algebra	math-word-problem	Yes	Yes	aops_forum	false
A point $P$ in the interior of a convex polyhedron in Euclidean space is called a [i]pivot point[/i] of the polyhedron if every line through $P$ contains exactly $0$ or $2$ vertices of the polyhedron. Determine, with proof, the maximum number of pivot points that a polyhedron can contain.	1. Assume for contradiction that a convex polyhedron $\Pi$ has two or more pivot points, and let two of them be $P$ and $Q$. 2. Construct a graph $G$ where the vertices of $G$ represent the vertices of $\Pi$. Connect two vertices with a red edge if they are collinear with $P$ and a blue edge if they are collinear with $Q$. 3. By the definition of a pivot point, each vertex of $G$ has exactly one red edge and one blue edge incident to it. Therefore, each vertex of $G$ has degree 2. 4. Consider any two vertices $A$ and $B$ of $\Pi$ collinear with $P$. By the definition of a pivot point, $P$ lies on the interior of $\Pi$, so $A$, $P$, and $B$ are collinear in this order. Similarly, for any two vertices $A$ and $B$ of $\Pi$ collinear with $Q$, $A$, $Q$, and $B$ are collinear in this order. 5. For a vertex $V_1$ of $G$, consider the sequence of adjacent vertices $V_1, V_2, V_3, \dots$ such that $V_i \neq V_{i+2}$. Each successive term of this sequence is a vertex not contained previously in the sequence. Since there are a finite number of distinct vertices, this sequence must eventually return to $V_1$. Therefore, each vertex of $G$ is contained in a cycle. 6. Let $V_1, V_2, V_3, \dots, V_m$ be one such cycle with period $m$ containing a vertex $V_1$. Since each vertex has a red and blue edge incident to it, $m$ must be even. This follows from counting the total number of red edges: each vertex has exactly one red edge incident to it, and each red edge is counted twice from the two vertices connected to it, giving a total of $\frac{m}{2}$ red edges. If $m$ were odd, the number of red edges would not be an integer. 7. Consider a plane containing line $V_{m-1}V_m$ not containing any other vertices of $\Pi$. Such a plane exists because there are finitely many vertices of $\Pi$ and infinitely many angles at which we can form a plane containing $V_{m-1}V_m$. Call this plane the "blue plane". This plane divides space into two sides: the "left side" containing $V_1$ and the "right side". 8. We assert that any blue edge contains a vertex on each side of the blue plane. This follows from the fact that for any two vertices $A$ and $B$ of $\Pi$ collinear with $Q$, $A$, $Q$, and $B$ are collinear in this order. Therefore, $V_1$ is on the left side, $V_2$ is on the right side, $V_3$ is on the left side, $V_4$ is on the right side, and in general $V_{2i-1}$ is on the left side and $V_{2i}$ is on the right side, except for $V_{m-1}$ and $V_m$ which lie on the blue plane. 9. Since $V_{m-1}V_m$ is a blue edge, $V_{m-2}V_{m-1}$ is a red edge. Since $m$ is even, $V_{m-2}$ lies on the right side of the blue plane. $V_{m-2}$, $P$, and $V_{m-1}$ are collinear in that order. Thus, the line $V_{m-2}P$ intersects the blue plane at a point other than $V_{m-1}$, which is a contradiction because two lines can intersect only once. Therefore, a convex polyhedron cannot have two pivot points. 10. Now, we prove that $n=1$ works. Consider a cube (which is a convex polyhedron) with the pivot point as the center of the cube. By symmetry, each line connecting a vertex of the cube through the center of the cube also passes through the opposite vertex. Hence, the center is one pivot point. Thus, $n=\boxed{1}$ is our answer.	1	Geometry	math-word-problem	Yes	Yes	aops_forum	false
Find the smallest positive integer $n$ that satisfies the following: We can color each positive integer with one of $n$ colors such that the equation $w + 6x = 2y + 3z$ has no solutions in positive integers with all of $w, x, y$ and $z$ having the same color. (Note that $w, x, y$ and $z$ need not be distinct.)	1. Prove that $ n > 3 $: - Assume for the sake of contradiction that there are at most 3 colors, say $ c_1 $, $ c_2 $, and $ c_3 $. - Let 1 have color $ c_1 $ without loss of generality. - By considering the tuple $(1,1,2,1)$, the color of 2 must be different from $ c_1 $, so let it be $ c_2 $. - By considering the tuple $(3,2,3,3)$, the color of 3 cannot be $ c_2 $, and by $(3,1,3,1)$, the color of 3 cannot be $ c_1 $. Therefore, the color of 3 must be $ c_3 $. - By considering the tuple $(6,2,6,2)$, the color of 6 cannot be $ c_2 $, and by $(3,3,6,3)$, the color of 6 cannot be $ c_3 $. Therefore, the color of 6 must be $ c_1 $. - By considering the tuple $(9,6,9,9)$, the color of 9 cannot be $ c_1 $, and by $(9,3,9,3)$, the color of 9 cannot be $ c_3 $. Therefore, the color of 9 must be $ c_2 $. - By considering the tuple $(6,4,6,6)$, the color of 4 cannot be $ c_1 $, and by $(2,2,4,2)$, the color of 4 cannot be $ c_2 $. Therefore, the color of 4 must be $ c_3 $. - By considering the tuple $(6,6,12,6)$, the color of 12 cannot be $ c_1 $, and by $(12,4,12,4)$, the color of 12 cannot be $ c_3 $. Therefore, the color of 12 must be $ c_2 $. - However, we have reached a contradiction, as $(12,2,9,2)$ is colored with only $ c_2 $. Therefore, there must be more than 3 colors. 2. Prove that $ n = 4 $ works: - Define the following four sets: \[ \begin{align} c_1 &= \{3^{2a}(3b+1) \mid a, b \ge 0\} \\ c_2 &= \{3^{2a}(3b+2) \mid a, b \ge 0\} \\ c_3 &= \{3^{2a+1}(3b+1) \mid a, b \ge 0\} \\ c_4 &= \{3^{2a+1}(3b+2) \mid a, b \ge 0\} \end{align} \] - It is obvious that each positive integer appears in exactly one of these sets because if we divide out all powers of 3 in a number, then we will get a number that is either 1 or 2 modulo 3. - We assert that no quadruple of positive integers $(w, x, y, z)$ satisfying $ w + 6x = 2y + 3z $ consists of four members from the same set. - Assume for the sake of contradiction that $(a, b, c, d)$ are positive integers satisfying $ a + 6b = 2c + 3d $ with $ a, b, c, d $ from the same set. - If $(a, b, c, d)$ are from the same set, then $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}, \frac{d}{3}\right)$ are also from the same set. Hence, we may assume without loss of generality that at least one of $(a, b, c, d)$ is not divisible by 3, or else we can perform this operation to reduce it. - Let $ v_3(t) $ denote the largest integer such that $ 3^{v_3(t)} \mid t $. Since at least one of these is not divisible by 3, $ v_3(a), v_3(b), v_3(c), v_3(d) $ must be even. - Notice that $ w + 6x \equiv w \equiv 2y + 3z \equiv -y \pmod{3} $, so $ w + y \equiv 0 \pmod{3} $. Therefore, $ w \equiv y \equiv 0 \pmod{3} $, or else they would have different residues modulo 3. Also, they must both be divisible by 9 because $ v_3(w) $ and $ v_3(y) $ are even. - Since $ 2y - w = 3(2x - z) $, $ 3 \mid 2x - z \implies x + z \equiv 0 \pmod{3} $. However, this implies that they are both divisible by 3, or else they would have different residues modulo 3. This is a contradiction, as we have assumed that at least one of $ w, x, y, z $ is not divisible by 3. Therefore, this coloring works. Thus, $ n = \boxed{4} $ is our answer.	4	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Let $a_1,a_2,a_3,...$ be a sequence of positive real numbers such that: (i) For all positive integers $m,n$, we have $a_{mn}=a_ma_n$ (ii) There exists a positive real number $B$ such that for all positive integers $m,n$ with $m<n$, we have $a_m < Ba_n$ Find all possible values of $\log_{2015}(a_{2015}) - \log_{2014}(a_{2014})$	1. Given Conditions and Initial Observations: - We are given a sequence of positive real numbers $a_1, a_2, a_3, \ldots$ such that: \[ a_{mn} = a_m a_n \quad \text{for all positive integers } m, n \] \[ \exists B > 0 \text{ such that } a_m < B a_n \quad \text{for all positive integers } m, n \text{ with } m < n \] - We need to find the value of $\log_{2015}(a_{2015}) - \log_{2014}(a_{2014})$. 2. Sequence Property Analysis: - From the first condition, we can deduce that $a_{n^k} = a_n^k$ by repeated application of $a_{mn} = a_m a_n$. - From the second condition, if $m < n$, then: \[ B > \frac{a_{m^k}}{a_{n^k}} = \left(\frac{a_m}{a_n}\right)^k \] This implies that $\frac{a_m}{a_n} < B^{1/k}$ for any $k$. As $k \to \infty$, $B^{1/k} \to 1$, so $\frac{a_m}{a_n} \leq 1$. Hence, $a_m \leq a_n$, proving that the sequence is non-decreasing. 3. Determining $a_1$: - Since $a_{1 \cdot 1} = a_1 \cdot a_1$, we have $a_1 = a_1^2$. Given $a_1$ is positive, it follows that $a_1 = 1$. 4. General Form of $a_n$: - We assert that $a_n = n^c$ for some real number $c$. To verify this, consider the prime factorization of $n$: \[ n = \prod_{i} p_i^{e_i} \] Using the property $a_{mn} = a_m a_n$, we get: \[ a_n = a_{\prod_{i} p_i^{e_i}} = \prod_{i} a_{p_i^{e_i}} = \prod_{i} (a_{p_i})^{e_i} \] Let $a_{p_i} = p_i^c$. Then: \[ a_n = \prod_{i} (p_i^c)^{e_i} = \left(\prod_{i} p_i^{e_i}\right)^c = n^c \] 5. Verification of Conditions: - The form $a_n = n^c$ satisfies the first condition: \[ a_{mn} = (mn)^c = m^c n^c = a_m a_n \] - For the second condition, we need $a_m < B a_n$ for $m < n$: \[ \frac{a_m}{a_n} = \left(\frac{m}{n}\right)^c < B \] This holds for any $c \geq 0$. 6. Calculating the Desired Expression: - Given $a_n = n^c$, we have: \[ \log_{2015}(a_{2015}) = \log_{2015}(2015^c) = c \] \[ \log_{2014}(a_{2014}) = \log_{2014}(2014^c) = c \] Therefore: \[ \log_{2015}(a_{2015}) - \log_{2014}(a_{2014}) = c - c = 0 \] The final answer is $\boxed{0}$	0	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Hello all. Post your solutions below. [b]Also, I think it is beneficial to everyone if you all attempt to comment on each other's solutions.[/b] 4/1/31. A group of $100$ friends stands in a circle. Initially, one person has $2019$ mangos, and no one else has mangos. The friends split the mangos according to the following rules: • sharing: to share, a friend passes two mangos to the left and one mango to the right. • eating: the mangos must also be eaten and enjoyed. However, no friend wants to be selfish and eat too many mangos. Every time a person eats a mango, they must also pass another mango to the right. A person may only share if they have at least three mangos, and they may only eat if they have at least two mangos. The friends continue sharing and eating, until so many mangos have been eaten that no one is able to share or eat anymore. Show that there are exactly eight people stuck with mangos, which can no longer be shared or eaten.	1. Initial Setup and Problem Understanding: - We have 100 friends standing in a circle. - Initially, one person has 2019 mangos, and no one else has any mangos. - Sharing rules: pass 2 mangos to the left and 1 mango to the right. - Eating rules: eat 1 mango and pass another mango to the right. - A person can only share if they have at least 3 mangos and can only eat if they have at least 2 mangos. - The goal is to show that there are exactly 8 people stuck with mangos that can no longer be shared or eaten. 2. Binary Representation Insight: - To solve this problem, we use the binary representation of the number of mangos. - The binary representation of 2019 is $2019_{10} = 11111100011_2$. - The number of 1s in the binary representation of 2019 is 8. 3. Binary Value Invariance: - We define the binary value of a distribution of mangos as the sum of the product of the number of mangos and the binary place value for all the friends. - We need to prove that neither sharing nor eating changes the binary value of a distribution. 4. Proof for Eating: - Consider three friends $A, B, C$ with binary place values $2^{x-1}, 2^x, 2^{x+1}$ respectively. - Let $a, b, c$ be the number of mangos they have. - The binary value of the distribution is $2^{x-1}a + 2^xb + 2^{x+1}c$. - If $b \geq 2$, friend $B$ can eat. After eating, $A$ still has $a$ mangos, $B$ has $b-2$ mangos, and $C$ has $c+1$ mangos. - The new binary value is: \[ 2^{x-1}a + 2^x(b-2) + 2^{x+1}(c+1) = 2^{x-1}a + 2^xb + 2^{x+1}c - 2(2^x) + 2^{x+1} \] - Since $-2(2^x) + 2^{x+1} = 0$, the binary value remains unchanged. 5. Proof for Sharing: - If $b \geq 3$, friend $B$ can share. After sharing, $A$ has $a+2$ mangos, $B$ has $b-3$ mangos, and $C$ has $c+1$ mangos. - The new binary value is: \[ 2^{x-1}(a+2) + 2^x(b-3) + 2^{x+1}(c+1) = 2^{x-1}a + 2^xb + 2^{x+1}c + 2(2^{x-1}) - 3(2^x) + 2^{x+1} \] - Since $2(2^{x-1}) + 2^{x+1} = 2^x + 2(2^x) = 3(2^x)$, the binary value remains unchanged. 6. Final Distribution: - Since neither sharing nor eating changes the binary value, the final distribution must have the same binary value as the initial distribution. - The binary value of the initial distribution is 2019. - The final distribution must correspond to the binary representation of 2019, which has 8 ones. 7. Conclusion: - Therefore, there are exactly 8 people stuck with mangos that can no longer be shared or eaten. The final answer is $\boxed{8}$	8	Logic and Puzzles	proof	Yes	Yes	aops_forum	false
Given a nonconstant polynomial with real coefficients $f(x),$ let $S(f)$ denote the sum of its roots. Let p and q be nonconstant polynomials with real coefficients such that $S(p) = 7,$ $S(q) = 9,$ and $S(p-q)= 11$. Find, with proof, all possible values for $S(p + q)$.	To solve the problem, we need to find the possible values of $ S(p + q) $ given that $ S(p) = 7 $, $ S(q) = 9 $, and $ S(p - q) = 11 $. We will use Vieta's formulas and properties of polynomial roots to derive the solution. 1. Expressing the sums of roots using Vieta's formulas: - For a polynomial $ p(x) $ of degree $ n $ with roots $ r_1, r_2, \ldots, r_n $, Vieta's formulas give us: \[ S(p) = r_1 + r_2 + \cdots + r_n = -\frac{a_{n-1}}{a_n} \] - Similarly, for a polynomial $ q(x) $ of degree $ m $ with roots $ s_1, s_2, \ldots, s_m $, we have: \[ S(q) = s_1 + s_2 + \cdots + s_m = -\frac{b_{m-1}}{b_m} \] - For the polynomial $ p(x) - q(x) $, the sum of the roots is: \[ S(p - q) = t_1 + t_2 + \cdots + t_k = -\frac{c_{k-1}}{c_k} \] 2. Given conditions: - $ S(p) = 7 $ - $ S(q) = 9 $ - $ S(p - q) = 11 $ 3. Analyzing the sum of roots of $ p(x) + q(x) $: - Let $ p(x) $ and $ q(x) $ be polynomials of degrees $ n $ and $ m $ respectively. - The polynomial $ p(x) + q(x) $ will have roots that are the sums of the roots of $ p(x) $ and $ q(x) $. 4. Using the given sums of roots: - We know that: \[ S(p) = 7, \quad S(q) = 9, \quad S(p - q) = 11 \] - We need to find $ S(p + q) $. 5. Using the properties of polynomial roots: - Consider the polynomial $ p(x) $ and $ q(x) $ with the given sums of roots. - The sum of the roots of $ p(x) + q(x) $ can be expressed as: \[ S(p + q) = S(p) + S(q) - S(p - q) \] 6. Calculating $ S(p + q) $: - Substitute the given values: \[ S(p + q) = 7 + 9 - 11 = 5 \] Therefore, the only possible value for $ S(p + q) $ is $ \boxed{5} $.	5	Algebra	math-word-problem	Yes	Yes	aops_forum	false
Find, with proof, all positive integers $n$ with the following property: There are only finitely many positive multiples of $n$ which have exactly $n$ positive divisors	1. Define the problem and notation: - We need to find all positive integers $ n $ such that there are only finitely many positive multiples of $ n $ which have exactly $ n $ positive divisors. - Let $\sigma(n)$ denote the number of positive divisors of $ n $. - Let $\mathbb{P}$ be the set of all primes. - Let $ D_n $ denote the set of prime divisors of $ n $. 2. Lemma 1: - For any positive integers $ n $ and $ k $, $\sigma(kn) \geq \sigma(n)$, with equality occurring only when $ k = 1 $. - Proof: - If $ k = 1 $, the lemma is trivially true. - Suppose $ k > 1 $. Any divisor of $ n $ must also be a divisor of $ kn $, but $ kn $ itself is a divisor of $ kn $ that does not divide $ n $. Thus, $\sigma(kn) > \sigma(n)$ for $ k > 1 $. - $\blacksquare$ 3. Lemma 2: - Given fixed nonnegative integers $ x_1, x_2, \ldots, x_k $ and a positive integer $ n $, there are finitely many ordered tuples of positive integers $ (y_1, y_2, \ldots, y_k) $ which satisfy $\prod_{i=1}^k (x_i + y_i) = n$. - Proof: - Since everything being multiplied is a positive integer, we can establish the upper bound $ y_i \leq n $ for all $ i $, giving an upper bound of $ n^k $ possible ordered tuples. - Since $ n^k $ is finite, the number of such tuples is finite. - $\blacksquare$ 4. Lemma 3: - The given condition that there are only finitely many positive multiples of $ n $ which have exactly $ n $ positive divisors is satisfied if and only if there does not exist a positive integer $ k $ such that $\sigma(kn) \mid n$ and $\sigma(kn) \neq n$. - Proof: - If part: - Suppose there exists a positive integer $ k $ such that $\sigma(kn) \mid n$ and $\sigma(kn) \neq n$. - Let $ m = \frac{n}{\sigma(kn)} $. Then $ m $ is a positive integer at least 2. - Every number of the form $ knp^{m-1} $ where $ p \in \mathbb{P} \setminus D_{kn} $ has $ n $ divisors. - Since $ \|\mathbb{P}\| $ is infinite and $ \|D_{kn}\| $ is finite, there are infinitely many such multiples. - Thus, if such a $ k $ exists, the condition is not satisfied. - Only if part: - Suppose there does not exist a positive integer $ k $ such that $\sigma(kn) \mid n$ and $\sigma(kn) \neq n$. - Any positive integer $ k $ such that $\sigma(kn) = n$ must satisfy $ D_k \subseteq D_n $. - Otherwise, there would exist a prime $ q \in D_k \setminus D_n $, leading to a contradiction. - Thus, any $ k $ such that $\sigma(kn) = n$ satisfies $ D_{kn} = D_n $. - Using the formula for $\sigma(kn)$, there are finitely many ordered tuples $(b_1, b_2, \ldots, b_m)$ such that $\prod_{i=1}^m (a_i + b_i + 1) = n$, by Lemma 2. - $\blacksquare$ 5. Positive integers not divisible by the square of any prime: - For $ n = 1 $, the only positive integer with one divisor is 1. - For $ n \neq 1 $, write $ n = \prod_{i=1}^m p_i $ for distinct primes $ p_i $. - Any $ k $ such that $\sigma(kn) = n$ must satisfy $ D_k \subseteq D_n $. - Using the formula for $\sigma(kn)$, there are finitely many ordered tuples $(a_1, a_2, \ldots, a_m)$ such that $\prod_{i=1}^m (a_i + 1) = n$, by Lemma 2. - Thus, any positive integers not divisible by the square of any prime work. 6. Special case for $ n = 4 $: - $\sigma(4) = 3$, so for any positive integer $ k $, $\sigma(4k) \geq 3$. - Thus, there does not exist a positive integer $ k $ such that $\sigma(4k) \mid 4$ and $\sigma(4k) \neq 4$. - By Lemma 3, there are only finitely many positive multiples of 4 which have exactly 4 divisors. 7. Induction for other cases: - Base cases: - For $ 2^k $ with $ k \geq 3 $, $\sigma(2^k) = k + 1 \leq 2^{k-1}$. - For $ p^k $ with $ k \geq 2 $ and $ p \in \mathbb{P} \setminus \{2\} $, $\sigma(p^k) = k + 1 \leq p^{k-1}$. - Inductive step: - Suppose $ n = \prod_{i=1}^m p_i^{a_i} $ doesn't work. - There exists a positive integer $ k $ such that $\sigma(kn) \mid n$ and $\sigma(kn) \neq n$. - Multiplying $ n $ by $ q^b $ for $ q \in \mathbb{P} \setminus D_n $ and $ b \in \mathbb{Z}^+ $ leads to infinitely many multiples with $ n \cdot q^b $ divisors. - This completes the inductive step. The final answer is all positive integers which are not divisible by the square of any prime, as well as $ \boxed{ 4 } $.	4	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Malmer Pebane's apartment uses a six-digit access code, with leading zeros allowed. He noticed that his fingers leave that reveal which digits were pressed. He decided to change his access code to provide the largest number of possible combinations for a burglar to try when the digits are known. For each number of distinct digits that could be used in the access code, calculate the number of possible combinations when the digits are known but their order and frequency are not known. For example, if there are smudges on $3$ and $9,$ two possible codes are $393939$ and $993999.$ Which number of distinct digits in the access code offers the most combinations?	1. Define $ f(x) $ as the number of possible combinations for the access code given that it has $ x $ distinct digits, for all integers $ 1 \leq x \leq 6 $. Clearly, $ f(1) = 1 $. 2. We claim that $ f(n) = n^6 - \left( \sum_{i=1}^{n-1} \binom{n}{i} f(i) \right) $ for all integers $ 2 \leq n \leq 6 $. 3. To prove this, note that the number of ways to have a 6-digit access code using a given set of $ n $ digits is $ n^6 $ if not all of the digits must be used. We now proceed with complementary counting. 4. Since $ n $ is an integer $ 2 \leq n \leq 6 $, this means that $ n^6 $ equals the sum of the number of ways to have a 6-digit access code using $ a $ of the digits in the set of $ n $ digits, where $ a $ is any arbitrary positive integer less than or equal to $ n $. 5. When $ a = n $, we can only choose one set with $ a $ elements that is a subset of our original set of $ n $ digits, so the number of 6-digit access codes formable using all $ n $ of the digits in the set of $ n $ digits is $ f(n) $. 6. Thus, $ f(n) $ is $ n^6 $ minus the number of 6-digit access codes that can be formed by only using a proper subset of the original set of $ n $ digits. 7. Therefore, $ f(n) $ is equal to $ n^6 $ minus the number of six-digit access codes that can be formed by only using exactly $ b $ of the digits in the original set, where $ b $ ranges from the integers from 1 to $ n-1 $. 8. For any particular value of $ b $, there are $ \binom{n}{b} $ subsets of the set of $ n $ digits with $ b $ digits. Also, for any subset $ C $ of the set of $ n $ digits with $ b $ elements, the number of ways to form a 6-digit access code using all of the elements in $ C $ is $ f(b) $, so the number of ways to form a 6-digit access code using exactly $ b $ digits from the subset of the $ n $ digits is $ \binom{n}{b} f(b) $ for all integers $ 1 \leq b \leq n-1 $. 9. A 6-digit access code cannot be formed without using any digits, so the number of ways to form a 6-digit access code by only using digits from a proper subset of the set of $ n $ digits is $ \sum_{i=1}^{n-1} \binom{n}{i} f(i) $. 10. Thus, $ f(n) = n^6 - \left( \sum_{i=1}^{n-1} \binom{n}{i} f(i) \right) $ for all integers $ 2 \leq n \leq 6 $. 11. Now, we can calculate $ f(i) $ for all integers $ 2 \leq i \leq 6 $. \[ \begin{align} f(2) &= 2^6 - \left( \sum_{i=1}^{1} \binom{2}{i} f(i) \right) = 64 - 2 = 62, \\ f(3) &= 3^6 - \left( \sum_{i=1}^{2} \binom{3}{i} f(i) \right) = 729 - 3 \cdot 62 - 3 \cdot 1 = 729 - 189 = 540, \\ f(4) &= 4^6 - \left( \sum_{i=1}^{3} \binom{4}{i} f(i) \right) = 4096 - 4 \cdot 540 - 6 \cdot 62 - 4 \cdot 1 = 4096 - 2536 = 1560, \\ f(5) &= 5^6 - \left( \sum_{i=1}^{4} \binom{5}{i} f(i) \right) = 15625 - 5 \cdot 1560 - 10 \cdot 540 - 10 \cdot 62 - 5 \cdot 1 = 15625 - 13825 = 1800, \\ f(6) &= 6^6 - \left( \sum_{i=1}^{5} \binom{6}{i} f(i) \right) = 46656 - 6 \cdot 1800 - 15 \cdot 1560 - 20 \cdot 540 - 15 \cdot 62 - 6 \cdot 1 = 46656 - 45936 = 720. \end{align} \] 12. Therefore, $ f(1) = 1 $, $ f(2) = 62 $, $ f(3) = 540 $, $ f(4) = 1560 $, $ f(5) = 1800 $, and $ f(6) = 720 $. 13. Thus, having 5 distinct digits gives the highest amount of possible access codes. The final answer is $ \boxed{5} $.	5	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] Len's Spanish class has four tests in the first term. Len scores $72$, $81$, and $78$ on the first three tests. If Len wants to have an 80 average for the term, what is the minimum score he needs on the last test? [b]p2.[/b] In $1824$, the Electoral College had $261$ members. Andrew Jackson won $99$ Electoral College votes and John Quincy Adams won $84$ votes. A plurality occurs when no candidate has more than $50\%$ of the votes. Should a plurality occur, the vote goes to the House of Representatives to break the tie. How many more votes would Jackson have needed so that a plurality would not have occurred? [b]p3.[/b] $\frac12 + \frac16 + \frac{1}{12} + \frac{1}{20} + \frac{1}{30}= 1 - \frac{1}{n}$. Find $n$. [b]p4.[/b] How many ways are there to sit Samuel, Esun, Johnny, and Prat in a row of $4$ chairs if Prat and Johnny refuse to sit on an end? [b]p5.[/b] Find an ordered quadruple $(w, x, y, z)$ that satisfies the following: $$3^w + 3^x + 3^y = 3^z$$ where $w + x + y + z = 2017$. [b]p6.[/b] In rectangle $ABCD$, $E$ is the midpoint of $CD$. If $AB = 6$ inches and $AE = 6$ inches, what is the length of $AC$? [b]p7.[/b] Call an integer interesting if the integer is divisible by the sum of its digits. For example, $27$ is divisible by $2 + 7 = 9$, so $27$ is interesting. How many $2$-digit interesting integers are there? [b]p8.[/b] Let $a\#b = \frac{a^3-b^3}{a-b}$ . If $a, b, c$ are the roots of the polynomial $x^3 + 2x^2 + 3x + 4$, what is the value of $a\#b + b\#c + c\#a$? [b]p9.[/b] Akshay and Gowri are examining a strange chessboard. Suppose $3$ distinct rooks are placed into the following chessboard. Find the number of ways that one can place these rooks so that they don't attack each other. Note that two rooks are considered attacking each other if they are in the same row or the same column. [img]https://cdn.artofproblemsolving.com/attachments/f/1/70f7d68c44a7a69eb13ce12291c0600d11027c.png[/img] [b]p10.[/b] The Earth is a very large sphere. Richard and Allen have a large spherical model of Earth, and they would like to (for some strange reason) cut the sphere up with planar cuts. If each cut intersects the sphere, and Allen holds the sphere together so it does not fall apart after each cut, what is the maximum number of pieces the sphere can be cut into after $6$ cuts? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	8. Given $a, b, c$ are the roots of the polynomial $x^3 + 2x^2 + 3x + 4$, we need to find the value of $a\#b + b\#c + c\#a$, where: \[ a\#b = \frac{a^3 - b^3}{a - b} = a^2 + ab + b^2 \] Therefore: \[ a\#b + b\#c + c\#a = a^2 + ab + b^2 + b^2 + bc + c^2 + c^2 + ca + a^2 \] \[ = 2(a^2 + b^2 + c^2) + ab + bc + ca \] Using Vieta's formulas for the roots of the polynomial $x^3 + 2x^2 + 3x + 4$: \[ a + b + c = -2, \quad ab + bc + ca = 3, \quad abc = -4 \] We need $a^2 + b^2 + c^2$: \[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = (-2)^2 - 2(3) = 4 - 6 = -2 \] Therefore: \[ a\#b + b\#c + c\#a = 2(-2) + 3 = -4 + 3 = -1 \] The final answer is $ \boxed{-1} $.	-1	Algebra	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] The remainder of a number when divided by $7$ is $5$. If I multiply the number by $32$ and add $18$ to the product, what is the new remainder when divided by $7$? [b]p2.[/b] If a fair coin is flipped $15$ times, what is the probability that there are more heads than tails? [b]p3.[/b] Let $-\frac{\sqrt{p}}{q}$ be the smallest nonzero real number such that the reciprocal of the number is equal to the number minus the square root of the square of the number, where $p$ and $q$ are positive integers and $p$ is not divisible the square of any prime. Find $p + q$. [b]p4.[/b] Rachel likes to put fertilizers on her grass to help her grass grow. However, she has cows there as well, and they eat $3$ little fertilizer balls on average. If each ball is spherical with a radius of $4$, then the total volume that each cow consumes can be expressed in the form $a\pi$ where $a$ is an integer. What is $a$? [b]p5.[/b] One day, all $30$ students in Precalc class are bored, so they decide to play a game. Everyone enters into their calculators the expression $9 \diamondsuit 9 \diamondsuit 9 ... \diamondsuit 9$, where $9$ appears $2020$ times, and each $\diamondsuit$ is either a multiplication or division sign. Each student chooses the signs randomly, but they each choose one more multiplication sign than division sign. Then all $30$ students calculate their expression and take the class average. Find the expected value of the class average. [b]p6.[/b] NaNoWriMo, or National Novel Writing Month, is an event in November during which aspiring writers attempt to produce novel-length work - formally defined as $50,000$ words or more - within the span of $30$ days. Justin wants to participate in NaNoWriMo, but he's a busy high school student: after accounting for school, meals, showering, and other necessities, Justin only has six hours to do his homework and perhaps participate in NaNoWriMo on weekdays. On weekends, he has twelve hours on Saturday and only nine hours on Sunday, because he goes to church. Suppose Justin spends two hours on homework every single day, including the weekends. On Wednesdays, he has science team, which takes up another hour and a half of his time. On Fridays, he spends three hours in orchestra rehearsal. Assume that he spends all other time on writing. Then, if November $1$st is a Friday, let $w$ be the minimum number of words per minute that Justin must type to finish the novel. Round $w$ to the nearest whole number. [b]p7.[/b] Let positive reals $a$, $b$, $c$ be the side lengths of a triangle with area $2030$. Given $ab + bc + ca = 15000$ and $abc = 350000$, find the sum of the lengths of the altitudes of the triangle. [b]p8.[/b] Find the minimum possible area of a rectangle with integer sides such that a triangle with side lengths $3$, $4$, $5$, a triangle with side lengths $4$, $5$, $6$, and a triangle with side lengths $\frac94$, $4$, $4$ all fit inside the rectangle without overlapping. [b]p9.[/b] The base $16$ number $10111213...99_{16}$, which is a concatenation of all of the (base $10$) $2$-digit numbers, is written on the board. Then, the last $2n$ digits are erased such that the base $10$ value of remaining number is divisible by $51$. Find the smallest possible integer value of $n$. [b]p10.[/b] Consider sequences that consist entirely of $X$'s, $Y$ 's and $Z$'s where runs of consecutive $X$'s, $Y$ 's, and $Z$'s are at most length $3$. How many sequences with these properties of length $8$ are there? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. We start with the given equation involving the smallest nonzero real number $ x $ such that the reciprocal of the number is equal to the number minus the square root of the square of the number. Mathematically, this can be written as: \[ \frac{1}{x} = x - \sqrt{x^2} \] 2. Since $\sqrt{x^2} = \|x\|$, the equation becomes: \[ \frac{1}{x} = x - \|x\| \] 3. We need to consider the cases for $ x $: - If $ x > 0 $, then $\|x\| = x$, and the equation simplifies to: \[ \frac{1}{x} = x - x = 0 \] This is not possible since $\frac{1}{x}$ cannot be zero. - If $ x < 0 $, then $\|x\| = -x$, and the equation simplifies to: \[ \frac{1}{x} = x - (-x) = x + x = 2x \] Thus, we have: \[ \frac{1}{x} = 2x \] 4. Solving for $ x $, we multiply both sides by $ x $: \[ 1 = 2x^2 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \] 5. Since $ x < 0 $, we take the negative value: \[ x = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \] 6. We are given that $ x = -\frac{\sqrt{p}}{q} $. Comparing this with $ x = -\frac{\sqrt{2}}{2} $, we identify $ p = 2 $ and $ q = 2 $. 7. Finally, we need to find $ p + q $: \[ p + q = 2 + 2 = 4 \] The final answer is $ \boxed{4} $.	4	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] If $a \diamond b = ab - a + b$, find $(3 \diamond 4) \diamond 5$ [b]p2.[/b] If $5$ chickens lay $5$ eggs in $5$ days, how many chickens are needed to lay $10$ eggs in $10$ days? [b]p3.[/b] As Alissa left her house to go to work one hour away, she noticed that her odometer read $16261$ miles. This number is a "special" number for Alissa because it is a palindrome and it contains exactly $1$ prime digit. When she got home that evening, it had changed to the next greatest "special" number. What was Alissa's average speed, in miles per hour, during her two hour trip? [b]p4.[/b] How many $1$ in by $3$ in by $8$ in blocks can be placed in a $4$ in by $4$ in by $9$ in box? [b]p5.[/b] Apple loves eating bananas, but she prefers unripe ones. There are $12$ bananas in each bunch sold. Given any bunch, if there is a $\frac13$ probability that there are $4$ ripe bananas, a $\frac16$ probability that there are $6$ ripe bananas, and a $\frac12$ probability that there are $10$ ripe bananas, what is the expected number of unripe bananas in $12$ bunches of bananas? [b]p6.[/b] The sum of the digits of a $3$-digit number $n$ is equal to the same number without the hundreds digit. What is the tens digit of $n$? [b]p7.[/b] How many ordered pairs of positive integers $(a, b)$ satisfy $a \le 20$, $b \le 20$, $ab > 15$? [b]p8.[/b] Let $z(n)$ represent the number of trailing zeroes of $n!$. What is $z(z(6!))?$ (Note: $n! = n\cdot (n-1) \cdot\cdot\cdot 2 \cdot 1$) [b]p9.[/b] On the Cartesian plane, points $A = (-1, 3)$, $B = (1, 8)$, and $C = (0, 10)$ are marked. $\vartriangle ABC$ is reflected over the line $y = 2x + 3$ to obtain $\vartriangle A'B'C'$. The sum of the $x$-coordinates of the vertices of $\vartriangle A'B'C'$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$, $b$. Compute $a + b$. [b]p10.[/b] How many ways can Bill pick three distinct points from the figure so that the points form a non-degenerate triangle? [img]https://cdn.artofproblemsolving.com/attachments/6/a/8b06f70d474a071b75556823f70a2535317944.png[/img] [b]p11.[/b] Say piece $A$ is attacking piece $B$ if the piece $B$ is on a square that piece $A$ can move to. How many ways are there to place a king and a rook on an $8\times 8$ chessboard such that the rook isn't attacking the king, and the king isn't attacking the rook? Consider rotations of the board to be indistinguishable. (Note: rooks move horizontally or vertically by any number of squares, while kings move $1$ square adjacent horizontally, vertically, or diagonally). [b]p12.[/b] Let the remainder when $P(x) = x^{2020} - x^{2017} - 1$ is divided by $S(x) = x^3 - 7$ be the polynomial $R(x) = ax^2 + bx + c$ for integers $a$, $b$, $c$. Find the remainder when $R(1)$ is divided by $1000$. [b]p13.[/b] Let $S(x) = \left \lfloor \frac{2020}{x} \right\rfloor + \left \lfloor \frac{2020}{x + 1} \right\rfloor$. Find the number of distinct values $S(x)$ achieves for integers $x$ in the interval $[1, 2020]$. [b]p14.[/b] Triangle $\vartriangle ABC$ is inscribed in a circle with center $O$ and has sides $AB = 24$, $BC = 25$, $CA = 26$. Let $M$ be the midpoint of $\overline{AB}$. Points $K$ and $L$ are chosen on sides $\overline{BC}$ and $\overline{CA}$, respectively such that $BK < KC$ and $CL < LA$. Given that $OM = OL = OK$, the area of triangle $\vartriangle MLK$ can be expressed as $\frac{a\sqrt{b}}{c}$ where $a, b, c$ are positive integers, $gcd(a, c) = 1$ and $b$ is not divisible by the square of any prime. Find $a + b + c$. [b]p15.[/b] Euler's totient function, $\phi (n)$, is defined as the number of positive integers less than $n$ that are relatively prime to $n$. Let $S(n)$ be the set of composite divisors of $n$. Evaluate $$\sum^{50}_{k=1}\left( k - \sum_{d\in S(k)} \phi (d) \right)$$ PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To solve the problem, we need to find the number of trailing zeroes in $6!$ and then find the number of trailing zeroes in that result. 1. Calculate $z(6!)$: The number of trailing zeroes in $n!$ is given by the formula: \[ z(n!) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \cdots \] For $n = 6$: \[ z(6!) = \left\lfloor \frac{6}{5} \right\rfloor + \left\lfloor \frac{6}{25} \right\rfloor + \left\lfloor \frac{6}{125} \right\rfloor + \cdots \] Since $\left\lfloor \frac{6}{5} \right\rfloor = 1$ and all other terms are zero: \[ z(6!) = 1 \] 2. Calculate $z(z(6!))$: We need to find the number of trailing zeroes in $1!$: \[ z(1!) = \left\lfloor \frac{1}{5} \right\rfloor + \left\lfloor \frac{1}{25} \right\rfloor + \left\lfloor \frac{1}{125} \right\rfloor + \cdots \] Since all terms are zero: \[ z(1!) = 0 \] The final answer is $\boxed{0}$	0	Other	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] Let $X = 2022 + 022 + 22 + 2$. When $X$ is divided by $22$, there is a remainder of $R$. What is the value of $R$? [b]p2.[/b] When Amy makes paper airplanes, her airplanes fly $75\%$ of the time. If her airplane flies, there is a $\frac56$ chance that it won’t fly straight. Given that she makes $80$ airplanes, what is the expected number airplanes that will fly straight? [b]p3.[/b] It takes Joshua working alone $24$ minutes to build a birdhouse, and his son working alone takes $16$ minutes to build one. The effective rate at which they work together is the sum of their individual working rates. How long in seconds will it take them to make one birdhouse together? [b]p4.[/b] If Katherine’s school is located exactly $5$ miles southwest of her house, and her soccer tournament is located exactly $12$ miles northwest of her house, how long, in hours, will it take Katherine to bike to her tournament right after school given she bikes at $0.5$ miles per hour? Assume she takes the shortest path possible. [b]p5.[/b] What is the largest possible integer value of $n$ such that $\frac{4n+2022}{n+1}$ is an integer? [b]p6.[/b] A caterpillar wants to go from the park situated at $(8, 5)$ back home, located at $(4, 10)$. He wants to avoid routes through $(6, 7)$ and $(7, 10)$. How many possible routes are there if the caterpillar can move in the north and west directions, one unit at a time? [b]p7.[/b] Let $\vartriangle ABC$ be a triangle with $AB = 2\sqrt{13}$, $BC = 6\sqrt2$. Construct square $BCDE$ such that $\vartriangle ABC$ is not contained in square $BCDE$. Given that $ACDB$ is a trapezoid with parallel bases $\overline{AC}$, $\overline{BD}$, find $AC$. [b]p8.[/b] How many integers $a$ with $1 \le a \le 1000$ satisfy $2^a \equiv 1$ (mod $25$) and $3^a \equiv 1$ (mod $29$)? [b]p9.[/b] Let $\vartriangle ABC$ be a right triangle with right angle at $B$ and $AB < BC$. Construct rectangle $ADEC$ such that $\overline{AC}$,$\overline{DE}$ are opposite sides of the rectangle, and $B$ lies on $\overline{DE}$. Let $\overline{DC}$ intersect $\overline{AB}$ at $M$ and let $\overline{AE}$ intersect $\overline{BC}$ at $N$. Given $CN = 6$, $BN = 4$, find the $m+n$ if $MN^2$ can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m, n$. [b]p10.[/b] An elimination-style rock-paper-scissors tournament occurs with $16$ players. The $16$ players are all ranked from $1$ to $16$ based on their rock-paper-scissor abilities where $1$ is the best and $16$ is the worst. When a higher ranked player and a lower ranked player play a round, the higher ranked player always beats the lower ranked player and moves on to the next round of the tournament. If the initial order of players are arranged randomly, and the expected value of the rank of the $2$nd place player of the tournament can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m, n$ what is the value of $m+n$? [b]p11.[/b] Estimation (Tiebreaker) Estimate the number of twin primes (pairs of primes that differ by $2$) where both primes in the pair are less than $220022$. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. Determine the probability that an airplane flies straight: - The probability that an airplane flies is $0.75$ or $\frac{3}{4}$. - Given that an airplane flies, the probability that it won't fly straight is $\frac{5}{6}$. - Therefore, the probability that it flies straight is $1 - \frac{5}{6} = \frac{1}{6}$. 2. Calculate the combined probability: - The probability that an airplane both flies and flies straight is: \[ \frac{3}{4} \times \frac{1}{6} = \frac{3}{24} = \frac{1}{8} \] 3. Use the Linearity of Expectation to find the expected number of airplanes that will fly straight: - If Amy makes 80 airplanes, the expected number of airplanes that will fly straight is: \[ 80 \times \frac{1}{8} = 10 \] The final answer is $\boxed{10}$	10	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
[u]Round 1[/u] [b]1.1.[/b] A classroom has $29$ students. A teacher needs to split up the students into groups of at most $4$. What is the minimum number of groups needed? [b]1.2.[/b] On his history map quiz, Eric recalls that Sweden, Norway and Finland are adjacent countries, but he has forgotten which is which, so he labels them in random order. The probability that he labels all three countries correctly can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [b]1.3.[/b] In a class of $40$ sixth graders, the class average for their final test comes out to be $90$ (out of a $100$). However, a student brings up an issue with problem $5$, and $10$ students receive credit for this question, bringing the class average to a $90.75$. How many points was problem $5$ worth? [u]Round 2[/u] [b]2.1.[/b] Compute $1 - 2 + 3 - 4 + ... - 2022 + 2023$. [b]2.2.[/b] In triangle $ABC$, $\angle ABC = 75^o$. Point $D$ lies on side $AC$ such that $BD = CD$ and $\angle BDC$ is a right angle. Compute the measure of $\angle A$. [b]2.3.[/b] Joe is rolling three four-sided dice each labeled with positive integers from $1$ to $4$. The probability the sum of the numbers on the top faces of the dice is $6$ can be written as $\frac{p}{q}$ where $p$ and $q$ are relatively prime integers. Find $p + q$. [u]Round 3[/u] [b]3.1.[/b] For positive integers $a, b, c, d$ that satisfy $a + b + c + d = 23$, what is the maximum value of $abcd$? [b]3.2.[/b] A buckball league has twenty teams. Each of the twenty teams plays exactly five games with each of the other teams. If each game takes 1 hour and thirty minutes, then how many total hours are spent playing games? [b]3.3.[/b] For a triangle $\vartriangle ABC$, let $M, N, O$ be the midpoints of $AB$, $BC$, $AC$, respectively. Let $P, Q, R$ be points on $AB$, $BC$, $AC$ such that $AP =\frac13 AB$, $BQ =\frac13 BC$, and $CR =\frac13 AC$. The ratio of the areas of $\vartriangle MNO$ and $\vartriangle P QR$ can be expressed as $\frac{m}{n}$ , where $ m$ and $n$ are relatively prime positive integers. Find $m + n$. [u]Round 4[/u] [b]4.1.[/b] $2023$ has the special property that leaves a remainder of $1$ when divided by $2$, $21$ when divided by $22$, and $22$ when divided by $23$. Let $n$ equal the lowest integer greater than $2023$ with the above properties. What is $n$? [b]4.2.[/b] Ants $A, B$ are on points $(0, 0)$ and $(3, 3)$ respectively, and ant A is trying to get to $(3, 3)$ while ant $B$ is trying to get to $(0, 0)$. Every second, ant $A$ will either move up or right one with equal probability, and ant $B$ will move down or left one with equal probability. The probability that the ants will meet each other be $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find $a + b$. [b]4.3.[/b] Find the number of trailing zeros of $100!$ in base $ 49$. PS. You should use hide for answers. Rounds 5-9 have been posted [url=https://artofproblemsolving.com/community/c3h3129723p28347714]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. Problem 1.1: To find the minimum number of groups needed to split 29 students into groups of at most 4, we use the ceiling function to determine the smallest integer greater than or equal to the division of 29 by 4. \[ \left\lceil \frac{29}{4} \right\rceil = \left\lceil 7.25 \right\rceil = 8 \] Therefore, the minimum number of groups needed is $ \boxed{8} $. 2. Problem 1.2: Eric labels the three countries in random order. The total number of permutations of three countries is $3! = 6$. Only one of these permutations is correct. \[ \text{Probability} = \frac{1}{6} \] The fraction $\frac{1}{6}$ is already in simplest form, so $m = 1$ and $n = 6$. Thus, $m + n = 1 + 6 = \boxed{7}$. 3. Problem 1.3: The initial total score for the class is: \[ 90 \times 40 = 3600 \] After the adjustment, the new total score is: \[ 90.75 \times 40 = 3630 \] The increase in total score is: \[ 3630 - 3600 = 30 \] Since 10 students received extra credit, the points per student for problem 5 is: \[ \frac{30}{10} = 3 \] Therefore, problem 5 was worth $ \boxed{3} $ points. 4. Problem 2.1: We need to compute the alternating sum: \[ 1 - 2 + 3 - 4 + \ldots - 2022 + 2023 \] Grouping the terms in pairs: \[ (1 - 2) + (3 - 4) + \ldots + (2021 - 2022) + 2023 \] Each pair sums to $-1$, and there are 1011 such pairs: \[ -1 \times 1011 + 2023 = -1011 + 2023 = 1012 \] Therefore, the result is $ \boxed{1012} $. 5. Problem 2.2: Given $BD = CD$ and $\angle BDC = 90^\circ$, triangle $BDC$ is an isosceles right triangle. Thus, $\angle CBD = 45^\circ$. Since $\angle ABC = 75^\circ$: \[ \angle DBA = 75^\circ - 45^\circ = 30^\circ \] In triangle $ABD$, $\angle A$ is: \[ \angle A = 180^\circ - 30^\circ - 90^\circ = 60^\circ \] Therefore, $\angle A = \boxed{60^\circ}$. 6. Problem 2.3: The total number of possible outcomes when rolling three four-sided dice is: \[ 4^3 = 64 \] We need to find the number of outcomes where the sum is 6. The possible combinations are: - (4, 1, 1): 3 ways (since the 1s are indistinguishable) - (3, 2, 1): $3! = 6$ ways - (2, 2, 2): 1 way Total favorable outcomes: \[ 3 + 6 + 1 = 10 \] Probability: \[ \frac{10}{64} = \frac{5}{32} \] Therefore, $p = 5$ and $q = 32$, so $p + q = 5 + 32 = \boxed{37}$. 7. Problem 3.1: To maximize $abcd$ given $a + b + c + d = 23$, we want the values to be as close as possible. The optimal values are $a = 5$ and $b = c = d = 6$: \[ 5 \cdot 6 \cdot 6 \cdot 6 = 5 \cdot 216 = 1080 \] Therefore, the maximum value is $ \boxed{1080} $. 8. Problem 3.2: Each team plays 5 games with each of the other 19 teams. Total games: \[ \frac{20 \times 19 \times 5}{2} = 950 \] Each game takes 1.5 hours: \[ 950 \times 1.5 = 1425 \] Therefore, the total hours spent playing games is $ \boxed{1425} $. 9. Problem 3.3: Using barycentric coordinates, the areas of $\triangle MNO$ and $\triangle PQR$ are: \[ [MNO] = \frac{1}{4}[ABC], \quad [PQR] = \frac{1}{3}[ABC] \] Ratio: \[ \frac{[MNO]}{[PQR]} = \frac{\frac{1}{4}}{\frac{1}{3}} = \frac{3}{4} \] Therefore, $m = 3$ and $n = 4$, so $m + n = 3 + 4 = \boxed{7}$. 10. Problem 4.1: We need $n$ such that: \[ n \equiv -1 \pmod{2}, \quad n \equiv -1 \pmod{22}, \quad n \equiv -1 \pmod{23} \] This implies: \[ n \equiv -1 \pmod{506} \] The smallest $n > 2023$ is: \[ 2023 + 506 = 2529 \] Therefore, $n = \boxed{2529}$. 11. Problem 4.2: The probability that the ants meet after 3 moves: \[ (2^3)^2 = 8^2 = 64 \] Number of ways they can meet: \[ 2\binom{3}{0}^2 + 2\binom{3}{1}^2 = 2 \cdot 1^2 + 2 \cdot 3^2 = 2 + 18 = 20 \] Probability: \[ \frac{20}{64} = \frac{5}{16} \] Therefore, $a = 5$ and $b = 16$, so $a + b = 5 + 16 = \boxed{21}$. 12. Problem 4.3: To find the number of trailing zeros of $100!$ in base $49$: \[ \nu_{49}(100!) = \left\lfloor \frac{\nu_7(100!)}{2} \right\rfloor \] \[ \nu_7(100!) = \left\lfloor \frac{100}{7} \right\rfloor + \left\lfloor \frac{100}{49} \right\rfloor = 14 + 2 = 16 \] \[ \left\lfloor \frac{16}{2} \right\rfloor = 8 \] Therefore, the number of trailing zeros is $ \boxed{8} $.	8	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] It’s currently $6:00$ on a $12$ hour clock. What time will be shown on the clock $100$ hours from now? Express your answer in the form hh : mm. [b]p2.[/b] A tub originally contains $10$ gallons of water. Alex adds some water, increasing the amount of water by 20%. Barbara, unhappy with Alex’s decision, decides to remove $20\%$ of the water currently in the tub. How much water, in gallons, is left in the tub? Express your answer as an exact decimal. [b]p3.[/b] There are $2000$ math students and $4000$ CS students at Berkeley. If $5580$ students are either math students or CS students, then how many of them are studying both math and CS? [b]p4.[/b] Determine the smallest integer $x$ greater than $1$ such that $x^2$ is one more than a multiple of $7$. [b]p5.[/b] Find two positive integers $x, y$ greater than $1$ whose product equals the following sum: $$9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29.$$ Express your answer as an ordered pair $(x, y)$ with $x \le y$. [b]p6.[/b] The average walking speed of a cow is $5$ meters per hour. If it takes the cow an entire day to walk around the edges of a perfect square, then determine the area (in square meters) of this square. [b]p7.[/b] Consider the cube below. If the length of the diagonal $AB$ is $3\sqrt3$, determine the volume of the cube. [img]https://cdn.artofproblemsolving.com/attachments/4/d/3a6fdf587c12f2e4637a029f38444914e161ac.png[/img] [b]p8.[/b] I have $18$ socks in my drawer, $6$ colored red, $8$ colored blue and $4$ colored green. If I close my eyes and grab a bunch of socks, how many socks must I grab to guarantee there will be two pairs of matching socks? [b]p9.[/b] Define the operation $a @ b$ to be $3 + ab + a + 2b$. There exists a number $x$ such that $x @ b = 1$ for all $b$. Find $x$. [b]p10.[/b] Compute the units digit of $2017^{(2017^2)}$. [b]p11.[/b] The distinct rational numbers $-\sqrt{-x}$, $x$, and $-x$ form an arithmetic sequence in that order. Determine the value of $x$. [b]p12.[/b] Let $y = x^2 + bx + c$ be a quadratic function that has only one root. If $b$ is positive, find $\frac{b+2}{\sqrt{c}+1}$. [b]p13.[/b] Alice, Bob, and four other people sit themselves around a circular table. What is the probability that Alice does not sit to the left or right of Bob? [b]p14.[/b] Let $f(x) = \|x - 8\|$. Let $p$ be the sum of all the values of $x$ such that $f(f(f(x))) = 2$ and $q$ be the minimum solution to $f(f(f(x))) = 2$. Compute $p \cdot q$. [b]p15.[/b] Determine the total number of rectangles ($1 \times 1$, $1 \times 2$, $2 \times 2$, etc.) formed by the lines in the figure below: $ \begin{tabular}{ \| l \| c \| c \| r\| } \hline & & & \\ \hline & & & \\ \hline & & & \\ \hline & & & \\ \hline \end{tabular} $ [b]p16.[/b] Take a square $ABCD$ of side length $1$, and let $P$ be the midpoint of $AB$. Fold the square so that point $D$ touches $P$, and let the intersection of the bottom edge $DC$ with the right edge be $Q$. What is $BQ$? [img]https://cdn.artofproblemsolving.com/attachments/1/1/aeed2c501e34a40a8a786f6bb60922b614a36d.png[/img] [b]p17.[/b] Let $A$, $B$, and $k$ be integers, where $k$ is positive and the greatest common divisor of $A$, $B$, and $k$ is $1$. Define $x\# y$ by the formula $x\# y = \frac{Ax+By}{kxy}$ . If $8\# 4 = \frac12$ and $3\# 1 = \frac{13}{6}$ , determine the sum $A + B + k$. [b]p18.[/b] There are $20$ indistinguishable balls to be placed into bins $A$, $B$, $C$, $D$, and $E$. Each bin must have at least $2$ balls inside of it. How many ways can the balls be placed into the bins, if each ball must be placed in a bin? [b]p19.[/b] Let $T_i$ be a sequence of equilateral triangles such that (a) $T_1$ is an equilateral triangle with side length 1. (b) $T_{i+1}$ is inscribed in the circle inscribed in triangle $T_i$ for $i \ge 1$. Find $$\sum^{\infty}_{i=1} Area (T_i).$$ [b]p20.[/b] A [i]gorgeous [/i] sequence is a sequence of $1$’s and $0$’s such that there are no consecutive $1$’s. For instance, the set of all gorgeous sequences of length $3$ is $\{[1, 0, 0]$,$ [1, 0, 1]$, $[0, 1, 0]$, $[0, 0, 1]$, $[0, 0, 0]\}$. Determine the number of gorgeous sequences of length $7$. PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To solve the problem, we need to find the smallest integer $ x $ greater than 1 such that $ x^2 $ is one more than a multiple of 7. Mathematically, we need to find $ x $ such that: \[ x^2 \equiv 1 \pmod{7} \] We will check the squares of integers modulo 7 starting from $ x = 2 $. 1. Calculate $ 2^2 \mod 7 $: \[ 2^2 = 4 \quad \Rightarrow \quad 4 \mod 7 = 4 \] Since $ 4 \neq 1 $, $ x = 2 $ is not a solution. 2. Calculate $ 3^2 \mod 7 $: \[ 3^2 = 9 \quad \Rightarrow \quad 9 \mod 7 = 2 \] Since $ 2 \neq 1 $, $ x = 3 $ is not a solution. 3. Calculate $ 4^2 \mod 7 $: \[ 4^2 = 16 \quad \Rightarrow \quad 16 \mod 7 = 2 \] Since $ 2 \neq 1 $, $ x = 4 $ is not a solution. 4. Calculate $ 5^2 \mod 7 $: \[ 5^2 = 25 \quad \Rightarrow \quad 25 \mod 7 = 4 \] Since $ 4 \neq 1 $, $ x = 5 $ is not a solution. 5. Calculate $ 6^2 \mod 7 $: \[ 6^2 = 36 \quad \Rightarrow \quad 36 \mod 7 = 1 \] Since $ 1 = 1 $, $ x = 6 $ is a solution. Thus, the smallest integer $ x $ greater than 1 such that $ x^2 $ is one more than a multiple of 7 is $ x = 6 $. The final answer is $ \boxed{6} $.	6	Other	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] Consider a $4 \times 4$ lattice on the coordinate plane. At $(0,0)$ is Mori’s house, and at $(4,4)$ is Mori’s workplace. Every morning, Mori goes to work by choosing a path going up and right along the roads on the lattice. Recently, the intersection at $(2, 2)$ was closed. How many ways are there now for Mori to go to work? [b]p2.[/b] Given two integers, define an operation $$ such that if a and b are integers, then a $$ b is an integer. The operation $$ has the following properties: 1. $a a$ = 0 for all integers $a$. 2. $(ka + b) * a = b * a$ for integers $a, b, k$. 3. $0 \le b * a < a$. 4. If $0 \le b < a$, then $b * a = b$. Find $2017 * 16$. [b]p3.[/b] Let $ABC$ be a triangle with side lengths $AB = 13$, $BC = 14$, $CA = 15$. Let $A'$, $B'$, $C'$, be the midpoints of $BC$, $CA$, and $AB$, respectively. What is the ratio of the area of triangle $ABC$ to the area of triangle $A'B'C'$? [b]p4.[/b] In a strange world, each orange has a label, a number from $0$ to $10$ inclusive, and there are an infinite number of oranges of each label. Oranges with the same label are considered indistinguishable. Sally has 3 boxes, and randomly puts oranges in her boxes such that (a) If she puts an orange labelled a in a box (where a is any number from 0 to 10), she cannot put any other oranges labelled a in that box. (b) If any two boxes contain an orange that have the same labelling, the third box must also contain an orange with that labelling. (c) The three boxes collectively contain all types of oranges (oranges of any label). The number of possible ways Sally can put oranges in her $3$ boxes is $N$, which can be written as the product of primes: $$p_1^{e_1} p_2^{e_2}... p_k^{e_k}$$ where $p_1 \ne p_2 \ne p_3 ... \ne p_k$ and $p_i$ are all primes and $e_i$ are all positive integers. What is the sum $e_1 + e_2 + e_3 +...+ e_k$? [b]p5.[/b] Suppose I want to stack $2017$ identical boxes. After placing the first box, every subsequent box must either be placed on top of another one or begin a new stack to the right of the rightmost pile. How many different ways can I stack the boxes, if the order I stack them doesn’t matter? Express your answer as $$p_1^{e_1} p_2^{e_2}... p_n^{e_n}$$ where $p_1, p_2, p_3, ... , p_n$ are distinct primes and $e_i$ are all positive integers. PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	### p1 1. Total Paths Calculation: - A path from $(0,0)$ to $(4,4)$ can be represented as a sequence of 4 'U' (up) moves and 4 'R' (right) moves. - The total number of such sequences is given by the binomial coefficient: \[ \binom{8}{4} = \frac{8!}{4!4!} = 70 \] 2. Paths Through $(2,2)$: - To go from $(0,0)$ to $(2,2)$, we need 2 'U' and 2 'R' moves. The number of such sequences is: \[ \binom{4}{2} = \frac{4!}{2!2!} = 6 \] - Similarly, to go from $(2,2)$ to $(4,4)$, we again need 2 'U' and 2 'R' moves, which also gives: \[ \binom{4}{2} = 6 \] - Therefore, the total number of paths passing through $(2,2)$ is: \[ 6 \times 6 = 36 \] 3. Paths Avoiding $(2,2)$: - Subtract the number of paths passing through $(2,2)$ from the total number of paths: \[ 70 - 36 = 34 \] The final answer is $\boxed{1}$	1	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] If $x$ is a real number that satisfies $\frac{48}{x} = 16$, find the value of $x$. [b]p2.[/b] If $ABC$ is a right triangle with hypotenuse $BC$ such that $\angle ABC = 35^o$, what is $\angle BCA$ in degrees? [img]https://cdn.artofproblemsolving.com/attachments/a/b/0f83dc34fb7934281e0e3f988ac34f653cc3f1.png[/img] [b]p3.[/b] If $a\vartriangle b = a + b - ab$, find $4\vartriangle 9$. [b]p4.[/b] Grizzly is $6$ feet tall. He measures his shadow to be $4$ feet long. At the same time, his friend Panda helps him measure the shadow of a nearby lamp post, and it is $6$ feet long. How tall is the lamp post in feet? [b]p5.[/b] Jerry is currently twice as old as Tom was $7$ years ago. Tom is $6$ years younger than Jerry. How many years old is Tom? [b]p6.[/b] Out of the $10, 000$ possible four-digit passcodes on a phone, how many of them contain only prime digits? [b]p7.[/b] It started snowing, which means Moor needs to buy snow shoes for his $6$ cows and $7$ sky bison. A cow has $4$ legs, and a sky bison has $6$ legs. If Moor has 36 snow shoes already, how many more shoes does he need to buy? Assume cows and sky bison wear the same type of shoe and each leg gets one shoe. [b]p8.[/b] How many integers $n$ with $1 \le n \le 100$ have exactly $3$ positive divisors? [b]p9.[/b] James has three $3$ candies and $3$ green candies. $3$ people come in and each randomly take $2$ candies. What is the probability that no one got $2$ candies of the same color? Express your answer as a decimal or a fraction in lowest terms. [b]p10.[/b] When Box flips a strange coin, the coin can land heads, tails, or on the side. It has a $\frac{1}{10}$probability of landing on the side, and the probability of landing heads equals the probability of landing tails. If Box flips a strange coin $3$ times, what is the probability that the number of heads flipped is equal to the number of tails flipped? Express your answer as a decimal or a fraction in lowest terms. [b]p11.[/b] James is travelling on a river. His canoe goes $4$ miles per hour upstream and $6$ miles per hour downstream. He travels $8$ miles upstream and then $8$ miles downstream (to where he started). What is his average speed, in miles per hour? Express your answer as a decimal or a fraction in lowest terms. [b]p12.[/b] Four boxes of cookies and one bag of chips cost exactly $1000$ jelly beans. Five bags of chips and one box of cookies cost less than $1000$ jelly beans. If both chips and cookies cost a whole number of jelly beans, what is the maximum possible cost of a bag of chips? [b]p13.[/b] June is making a pumpkin pie, which takes the shape of a truncated cone, as shown below. The pie tin is $18$ inches wide at the top, $16$ inches wide at the bottom, and $1$ inch high. How many cubic inches of pumpkin filling are needed to fill the pie? [img]https://cdn.artofproblemsolving.com/attachments/7/0/22c38dd6bc42d15ad9352817b25143f0e4729b.png[/img] [b]p14.[/b] For two real numbers $a$ and $b$, let $a\# b = ab - 2a - 2b + 6$. Find a positive real number $x$ such that $(x\#7) \#x = 82$. [b]p15.[/b] Find the sum of all positive integers $n$ such that $\frac{n^2 + 20n + 51}{n^2 + 4n + 3}$ is an integer. [b]p16.[/b] Let $ABC$ be a right triangle with hypotenuse $AB$ such that $AC = 36$ and $BC = 15$. A semicircle is inscribed in $ABC$ as shown, such that the diameter $XC$ of the semicircle lies on side $AC$ and that the semicircle is tangent to $AB$. What is the radius of the semicircle? [img]https://cdn.artofproblemsolving.com/attachments/4/2/714f7dfd09f6da1d61a8f910b5052e60dcd2fb.png[/img] [b]p17.[/b] Let $a$ and $b$ be relatively prime positive integers such that the product $ab$ is equal to the least common multiple of $16500$ and $990$. If $\frac{16500}{a}$ and $\frac{990}{b}$ are both integers, what is the minimum value of $a + b$? [b]p18.[/b] Let $x$ be a positive real number so that $x - \frac{1}{x} = 1$. Compute $x^8 - \frac{1}{x^8}$ . [b]p19.[/b] Six people sit around a round table. Each person rolls a standard $6$-sided die. If no two people sitting next to each other rolled the same number, we will say that the roll is valid. How many dierent rolls are valid? [b]p20.[/b] Given that $\frac{1}{31} = 0.\overline{a_1a_2a_3a_4a_5... a_n}$ (that is, $\frac{1}{31}$ can be written as the repeating decimal expansion $0.a_1a_2... a_na_1a_2... a_na_1a_2...$ ), what is the minimum value of $n$? PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To solve the given problem, we need to find a positive real number $ x $ such that $(x \# 7) \# x = 82$, where the operation $\#$ is defined as $a \# b = ab - 2a - 2b + 6$. 1. First, we need to compute $ x \# 7 $: \[ x \# 7 = x \cdot 7 - 2x - 2 \cdot 7 + 6 = 7x - 2x - 14 + 6 = 5x - 8 \] 2. Next, we need to compute $(5x - 8) \# x$: \[ (5x - 8) \# x = (5x - 8) \cdot x - 2(5x - 8) - 2x + 6 \] Simplify the expression step-by-step: \[ (5x - 8) \cdot x = 5x^2 - 8x \] \[ -2(5x - 8) = -10x + 16 \] \[ -2x = -2x \] \[ \text{Combining all terms: } 5x^2 - 8x - 10x + 16 - 2x + 6 = 5x^2 - 20x + 22 \] 3. Set the expression equal to 82 and solve for $ x $: \[ 5x^2 - 20x + 22 = 82 \] \[ 5x^2 - 20x + 22 - 82 = 0 \] \[ 5x^2 - 20x - 60 = 0 \] 4. Solve the quadratic equation $ 5x^2 - 20x - 60 = 0 $: \[ x^2 - 4x - 12 = 0 \quad \text{(dividing the entire equation by 5)} \] \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, $ a = 1 $, $ b = -4 $, and $ c = -12 $: \[ x = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2} \] \[ x = \frac{4 + 8}{2} = 6 \quad \text{or} \quad x = \frac{4 - 8}{2} = -2 \] 5. Since we need a positive real number $ x $: \[ x = 6 \] The final answer is $ \boxed{6} $.	6	Algebra	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] What is the maximum number of points of intersection between a square and a triangle, assuming that no side of the triangle is parallel to any side of the square? [b]p2.[/b] Two angles of an isosceles triangle measure $80^o$ and $x^o$. What is the sum of all the possible values of $x$? [b]p3.[/b] Let $p$ and $q$ be prime numbers such that $p + q$ and p + $7q$ are both perfect squares. Find the value of $pq$. [b]p4.[/b] Anna, Betty, Carly, and Danielle are four pit bulls, each of which is either wearing or not wearing lipstick. The following three facts are true: (1) Anna is wearing lipstick if Betty is wearing lipstick. (2) Betty is wearing lipstick only if Carly is also wearing lipstick. (3) Carly is wearing lipstick if and only if Danielle is wearing lipstick The following five statements are each assigned a certain number of points: (a) Danielle is wearing lipstick if and only if Carly is wearing lipstick. (This statement is assigned $1$ point.) (b) If Anna is wearing lipstick, then Betty is wearing lipstick. (This statement is assigned $6$ points.) (c) If Betty is wearing lipstick, then both Anna and Danielle must be wearing lipstick. (This statement is assigned $10$ points.) (d) If Danielle is wearing lipstick, then Anna is wearing lipstick. (This statement is assigned $12$ points.) (e) If Betty is wearing lipstick, then Danielle is wearing lipstick. (This statement is assigned $14$ points.) What is the sum of the points assigned to the statements that must be true? (For example, if only statements (a) and (d) are true, then the answer would be $1 + 12 = 13$.) [b]p5.[/b] Let $f(x)$ and $g(x)$ be functions such that $f(x) = 4x + 3$ and $g(x) = \frac{x + 1}{4}$. Evaluate $g(f(g(f(42))))$. [b]p6.[/b] Let $A,B,C$, and $D$ be consecutive vertices of a regular polygon. If $\angle ACD = 120^o$, how many sides does the polygon have? [b]p7.[/b] Fred and George have a fair $8$-sided die with the numbers $0, 1, 2, 9, 2, 0, 1, 1$ written on the sides. If Fred and George each roll the die once, what is the probability that Fred rolls a larger number than George? [b]p8.[/b] Find the smallest positive integer $t$ such that $(23t)^3 - (20t)^3 - (3t)^3$ is a perfect square. [b]p9.[/b] In triangle $ABC$, $AC = 8$ and $AC < AB$. Point $D$ lies on side BC with $\angle BAD = \angle CAD$. Let $M$ be the midpoint of $BC$. The line passing through $M$ parallel to $AD$ intersects lines $AB$ and $AC$ at $F$ and $E$, respectively. If $EF =\sqrt2$ and $AF = 1$, what is the length of segment $BC$? (See the following diagram.) [img]https://cdn.artofproblemsolving.com/attachments/2/3/4b5dd0ae28b09f5289fb0e6c72c7cbf421d025.png[/img] [b]p10.[/b] There are $2011$ evenly spaced points marked on a circular table. Three segments are randomly drawn between pairs of these points such that no two segments share an endpoint on the circle. What is the probability that each of these segments intersects the other two? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. p1. What is the maximum number of points of intersection between a square and a triangle, assuming that no side of the triangle is parallel to any side of the square? To find the maximum number of points of intersection between a square and a triangle, we need to consider the following: - A square has 4 sides. - A triangle has 3 sides. Each side of the triangle can intersect each side of the square at most once. Therefore, the maximum number of intersections is given by the product of the number of sides of the square and the number of sides of the triangle: \[ 4 \times 3 = 12 \] Thus, the maximum number of points of intersection is $ \boxed{12} $.	12	Geometry	math-word-problem	Yes	Yes	aops_forum	false
[i]20 problems for 20 minutes.[/i] [b]p1.[/b] Euclid eats $\frac17$ of a pie in $7$ seconds. Euler eats $\frac15$ of an identical pie in $10$ seconds. Who eats faster? [b]p2.[/b] Given that $\pi = 3.1415926...$ , compute the circumference of a circle of radius 1. Express your answer as a decimal rounded to the nearest hundred thousandth (i.e. $1.234562$ and $1.234567$ would be rounded to $1.23456$ and $1.23457$, respectively). [b]p3.[/b] Alice bikes to Wonderland, which is $6$ miles from her house. Her bicycle has two wheels, and she also keeps a spare tire with her. If each of the three tires must be used for the same number of miles, for how many miles will each tire be used? [b]p4.[/b] Simplify $\frac{2010 \cdot 2010}{2011}$ to a mixed number. (For example, $2\frac12$ is a mixed number while $\frac52$ and $2.5$ are not.) [b]p5.[/b] There are currently $175$ problems submitted for $EMC^2$. Chris has submitted $51$ of them. If nobody else submits any more problems, how many more problems must Chris submit so that he has submitted $\frac13$ of the problems? [b]p6.[/b] As shown in the diagram below, points $D$ and $L$ are located on segment $AK$, with $D$ between $A$ and $L$, such that $\frac{AD}{DK}=\frac{1}{3}$ and $\frac{DL}{LK}=\frac{5}{9}$. What is $\frac{DL}{AK}$? [img]https://cdn.artofproblemsolving.com/attachments/9/a/3f92bd33ffbe52a735158f7ebca79c4c360d30.png[/img] [b]p7.[/b] Find the number of possible ways to order the letters $G, G, e, e, e$ such that two neighboring letters are never $G$ and $e$ in that order. [b]p8.[/b] Find the number of odd composite integers between $0$ and $50$. [b]p9.[/b] Bob tries to remember his $2$-digit extension number. He knows that the number is divisible by $5$ and that the first digit is odd. How many possibilities are there for this number? [b]p10.[/b] Al walks $1$ mile due north, then $2$ miles due east, then $3$ miles due south, and then $4$ miles due west. How far, in miles, is he from his starting position? (Assume that the Earth is flat.) [b]p11.[/b] When n is a positive integer, $n!$ denotes the product of the first $n$ positive integers; that is, $n! = 1 \cdot 2 \cdot 3 \cdot ... \cdot n$. Given that $7! = 5040$, compute $8! + 9! + 10!$. [b]p12.[/b] Sam's phone company charges him a per-minute charge as well as a connection fee (which is the same for every call) every time he makes a phone call. If Sam was charged $\$4.88$ for an $11$-minute call and $\$6.00$ for a $19$-minute call, how much would he be charged for a $15$-minute call? [b]p13.[/b] For a positive integer $n$, let $s_n$ be the sum of the n smallest primes. Find the least $n$ such that $s_n$ is a perfect square (the square of an integer). [b]p14.[/b] Find the remainder when $2011^{2011}$ is divided by $7$. [b]p15.[/b] Let $a, b, c$, and $d$ be $4$ positive integers, each of which is less than $10$, and let $e$ be their least common multiple. Find the maximum possible value of $e$. [b]p16.[/b] Evaluate $100 - 1 + 99 - 2 + 98 - 3 + ... + 52 - 49 + 51 - 50$. [b]p17.[/b] There are $30$ basketball teams in the Phillips Exeter Dorm Basketball League. In how ways can $4$ teams be chosen for a tournament if the two teams Soule Internationals and Abbot United cannot be chosen at the same time? [b]p18.[/b] The numbers $1, 2, 3, 4, 5, 6$ are randomly written around a circle. What is the probability that there are four neighboring numbers such that the sum of the middle two numbers is less than the sum of the other two? [b]p19.[/b] What is the largest positive $2$-digit factor of $3^{2^{2011}} - 2^{2^{2011}}$? [b]p20.[/b] Rhombus $ABCD$ has vertices $A = (-12,-4)$, $B = (6, b)$, $C = (c,-4)$ and $D = (d,-28)$, where $b$, $c$, and $d$ are integers. Find a constant $m$ such that the line y = $mx$ divides the rhombus into two regions of equal area. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To find the number of odd composite integers between 0 and 50, we need to follow these steps: 1. Identify the odd numbers between 0 and 50: The odd numbers between 0 and 50 are: \[ 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49 \] There are 25 odd numbers in this range. 2. Identify the odd prime numbers between 0 and 50: The odd prime numbers between 0 and 50 are: \[ 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 \] There are 14 odd prime numbers in this range. 3. Calculate the number of odd composite numbers: Odd composite numbers are odd numbers that are not prime. To find the number of odd composite numbers, subtract the number of odd primes from the total number of odd numbers: \[ 25 - 14 = 11 \] 4. List the odd composite numbers to verify: The odd composite numbers between 0 and 50 are: \[ 9, 15, 21, 25, 27, 33, 35, 39, 45, 49 \] There are indeed 10 odd composite numbers. The final answer is $\boxed{10}$.	10	Other	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] Determine the number of ways to place $4$ rooks on a $4 \times 4$ chessboard such that: (a) no two rooks attack one another, and (b) the main diagonal (the set of squares marked $X$ below) does not contain any rooks. [img]https://cdn.artofproblemsolving.com/attachments/e/e/e3aa96de6c8ed468c6ef3837e66a0bce360d36.png[/img] The rooks are indistinguishable and the board cannot be rotated. (Two rooks attack each other if they are in the same row or column.) [b]p2.[/b] Seven students, numbered $1$ to $7$ in counter-clockwise order, are seated in a circle. Fresh Mann has 100 erasers, and he wants to distribute them to the students, albeit unfairly. Starting with person $ 1$ and proceeding counter-clockwise, Fresh Mann gives $i$ erasers to student $i$; for example, he gives $ 1$ eraser to student $ 1$, then $2$ erasers to student $2$, et cetera. He continues around the circle until he does not have enough erasers to give to the next person. At this point, determine the number of erasers that Fresh Mann has. [b]p3.[/b] Let $ABC$ be a triangle with $AB = AC = 17$ and $BC = 24$. Approximate $\angle ABC$ to the nearest multiple of $10$ degrees. [b]p4.[/b] Define a sequence of rational numbers $\{x_n\}$ by $x_1 =\frac35$ and for $n \ge 1$, $x_{n+1} = 2 - \frac{1}{x_n}$ . Compute the product $x_1x_2x_3... x_{2013}$. [b]p5.[/b] In equilateral triangle $ABC$, points $P$ and $R$ lie on segment $AB$, points $I$ and $M$ lie on segment $BC$, and points $E$ and $S$ lie on segment $CA$ such that $PRIMES$ is a equiangular hexagon. Given that $AB = 11$, $PR = 2$, $IM = 3$, and $ES = 5$, compute the area of hexagon $PRIMES$. [b]p6.[/b] Let $f(a, b) = \frac{a^2}{a+b}$ . Let $A$ denote the sum of $f(i, j)$ over all pairs of integers $(i, j)$ with $1 \le i < j \le 10$; that is, $$A = (f(1, 2) + f(1, 3) + ...+ f(1, 10)) + (f(2, 3) + f(2, 4) +... + f(2, 10)) +... + f(9, 10).$$ Similarly, let $B$ denote the sum of $f(i, j)$ over all pairs of integers $(i, j)$ with $1 \le j < i \le 10$, that is, $$B = (f(2, 1) + f(3, 1) + ... + f(10, 1)) + (f(3, 2) + f(4, 2) +... + f(10, 2)) +... + f(10, 9).$$ Compute $B - A$. [b]p7.[/b] Fresh Mann has a pile of seven rocks with weights $1, 1, 2, 4, 8, 16$, and $32$ pounds and some integer X between $1$ and $64$, inclusive. He would like to choose a set of the rocks whose total weight is exactly $X$ pounds. Given that he can do so in more than one way, determine the sum of all possible values of $X$. (The two $1$-pound rocks are indistinguishable.) [b]p8.[/b] Let $ABCD$ be a convex quadrilateral with $AB = BC = CA$. Suppose that point $P$ lies inside the quadrilateral with $AP = PD = DA$ and $\angle PCD = 30^o$. Given that $CP = 2$ and $CD = 3$, compute $CA$. [b]p9.[/b] Define a sequence of rational numbers $\{x_n\}$ by $x_1 = 2$, $x_2 = \frac{13}{2}$ , and for $n \ge 1$, $x_{n+2} = 3 -\frac{3}{x_{n+1}}+\frac{1}{x_nx_{n+1}}$. Compute $x_{100}$. [b]p10.[/b] Ten prisoners are standing in a line. A prison guard wants to place a hat on each prisoner. He has two colors of hats, red and blue, and he has $10$ hats of each color. Determine the number of ways in which the prison guard can place hats such that among any set of consecutive prisoners, the number of prisoners with red hats and the number of prisoners with blue hats differ by at most $2$. PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. First, we need to determine the total number of ways to place 4 rooks on a $4 \times 4$ chessboard such that no two rooks attack each other. This is equivalent to finding the number of permutations of 4 elements, which is given by $4!$: \[ 4! = 24 \] 2. Next, we need to use the Principle of Inclusion/Exclusion (PIE) to count the number of ways to place the rooks such that at least one rook is on the main diagonal. The main diagonal consists of the squares $(1,1)$, $(2,2)$, $(3,3)$, and $(4,4)$. 3. Let $A_i$ be the set of arrangements where the $i$-th rook is on the $i$-th diagonal square. We need to find $\|A_1 \cup A_2 \cup A_3 \cup A_4\|$. 4. By PIE, we have: \[ \|A_1 \cup A_2 \cup A_3 \cup A_4\| = \sum_{i=1}^4 \|A_i\| - \sum_{1 \leq i < j \leq 4} \|A_i \cap A_j\| + \sum_{1 \leq i < j < k \leq 4} \|A_i \cap A_j \cap A_k\| - \|A_1 \cap A_2 \cap A_3 \cap A_4\| \] 5. Calculate each term: - $\|A_i\| = 3!$ because if one rook is fixed on the diagonal, the remaining 3 rooks can be placed in $3!$ ways: \[ \|A_i\| = 6 \] - $\|A_i \cap A_j\| = 2!$ because if two rooks are fixed on the diagonal, the remaining 2 rooks can be placed in $2!$ ways: \[ \|A_i \cap A_j\| = 2 \] - $\|A_i \cap A_j \cap A_k\| = 1!$ because if three rooks are fixed on the diagonal, the remaining rook can be placed in $1!$ way: \[ \|A_i \cap A_j \cap A_k\| = 1 \] - $\|A_1 \cap A_2 \cap A_3 \cap A_4\| = 0!$ because if all four rooks are fixed on the diagonal, there is only one way to place them: \[ \|A_1 \cap A_2 \cap A_3 \cap A_4\| = 1 \] 6. Substitute these values into the PIE formula: \[ \|A_1 \cup A_2 \cup A_3 \cup A_4\| = 4 \cdot 6 - \binom{4}{2} \cdot 2 + \binom{4}{3} \cdot 1 - 1 = 24 - 6 \cdot 2 + 4 \cdot 1 - 1 = 24 - 12 + 4 - 1 = 15 \] 7. Finally, subtract the number of invalid arrangements from the total number of arrangements: \[ 24 - 15 = 9 \] Conclusion: \[ \boxed{9} \]	9	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
[u]Round 5[/u] [b]p13.[/b] Five different schools are competing in a tournament where each pair of teams plays at most once. Four pairs of teams are randomly selected and play against each other. After these four matches, what is the probability that Chad's and Jordan's respective schools have played against each other, assuming that Chad and Jordan come from different schools? [b]p14.[/b] A square of side length $1$ and a regular hexagon are both circumscribed by the same circle. What is the side length of the hexagon? [b]p15.[/b] From the list of integers $1,2, 3,...,30$ Jordan can pick at least one pair of distinct numbers such that none of the $28$ other numbers are equal to the sum or the difference of this pair. Of all possible such pairs, Jordan chooses the pair with the least sum. Which two numbers does Jordan pick? [u]Round 6[/u] [b]p16.[/b] What is the sum of all two-digit integers with no digit greater than four whose squares also have no digit greater than four? [b]p17.[/b] Chad marks off ten points on a circle. Then, Jordan draws five chords under the following constraints: $\bullet$ Each of the ten points is on exactly one chord. $\bullet$ No two chords intersect. $\bullet$ There do not exist (potentially non-consecutive) points $A, B,C,D,E$, and $F$, in that order around the circle, for which $AB$, $CD$, and $EF$ are all drawn chords. In how many ways can Jordan draw these chords? [b]p18.[/b] Chad is thirsty. He has $109$ cubic centimeters of silicon and a 3D printer with which he can print a cup to drink water in. He wants a silicon cup whose exterior is cubical, with five square faces and an open top, that can hold exactly $234$ cubic centimeters of water when filled to the rim in a rectangular-box-shaped cavity. Using all of his silicon, he prints a such cup whose thickness is the same on the five faces. What is this thickness, in centimeters? [u]Round 7[/u] [b]p19.[/b] Jordan wants to create an equiangular octagon whose side lengths are exactly the first $8$ positive integers, so that each side has a different length. How many such octagons can Jordan create? [b]p20.[/b] There are two positive integers on the blackboard. Chad computes the sum of these two numbers and tells it to Jordan. Jordan then calculates the sum of the greatest common divisor and the least common multiple of the two numbers, and discovers that her result is exactly $3$ times as large as the number Chad told her. What is the smallest possible sum that Chad could have said? [b]p21.[/b] Chad uses yater to measure distances, and knows the conversion factor from yaters to meters precisely. When Jordan asks Chad to convert yaters into meters, Chad only gives Jordan the result rounded to the nearest integer meters. At Jordan's request, Chad converts $5$ yaters into $8$ meters and $7$ yaters into $12$ meters. Given this information, how many possible numbers of meters could Jordan receive from Chad when requesting to convert $2014$ yaters into meters? [u]Round 8[/u] [b]p22.[/b] Jordan places a rectangle inside a triangle with side lengths $13$, $14$, and $15$ so that the vertices of the rectangle all lie on sides of the triangle. What is the maximum possible area of Jordan's rectangle? [b]p23.[/b] Hoping to join Chad and Jordan in the Exeter Space Station, there are $2014$ prospective astronauts of various nationalities. It is given that $1006$ of the astronaut applicants are American and that there are a total of $64$ countries represented among the applicants. The applicants are to group into $1007$ pairs with no pair consisting of two applicants of the same nationality. Over all possible distributions of nationalities, what is the maximum number of possible ways to make the $1007$ pairs of applicants? Express your answer in the form $a \cdot b!$, where $a$ and $b$ are positive integers and $a$ is not divisible by $b + 1$. Note: The expression $k!$ denotes the product $k \cdot (k - 1) \cdot ... \cdot 2 \cdot 1$. [b]p24.[/b] We say a polynomial $P$ in $x$ and $y$ is $n$-[i]good [/i] if $P(x, y) = 0$ for all integers $x$ and $y$, with $x \ne y$, between $1$ and $n$, inclusive. We also define the complexity of a polynomial to be the maximum sum of exponents of $x$ and $y$ across its terms with nonzero coeffcients. What is the minimal complexity of a nonzero $4$-good polynomial? In addition, give an example of a $4$-good polynomial attaining this minimal complexity. PS. You should use hide for answers. Rounds 1-4 have been posted [url=https://artofproblemsolving.com/community/c3h2915803p26040550]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. Let the two positive integers on the blackboard be $a$ and $b$, with the greatest common divisor $d$. We can express $a$ and $b$ as: \[ a = da', \quad b = db' \] where $a'$ and $b'$ are coprime integers. 2. The sum of the greatest common divisor and the least common multiple of $a$ and $b$ is given by: \[ \gcd(a, b) + \text{lcm}(a, b) = d + \frac{ab}{d} \] Substituting $a = da'$ and $b = db'$, we get: \[ d + \frac{da' \cdot db'}{d} = d + d a' b' \] Simplifying, we have: \[ d + d a' b' = d(1 + a' b') \] 3. According to the problem, this sum is three times the sum of $a$ and $b$: \[ d(1 + a' b') = 3(a + b) = 3(da' + db') = 3d(a' + b') \] Dividing both sides by $d$, we obtain: \[ 1 + a' b' = 3(a' + b') \] 4. Rearranging the equation, we get: \[ 1 + a' b' = 3a' + 3b' \implies a' b' - 3a' - 3b' = -1 \implies (a' - 3)(b' - 3) = 8 \] 5. We need to find pairs $(a', b')$ such that $(a' - 3)(b' - 3) = 8$. The integer solutions to this equation are: \[ (a' - 3, b' - 3) = (1, 8), (8, 1), (2, 4), (4, 2) \] Thus, the pairs $(a', b')$ are: \[ (a', b') = (4, 11), (11, 4), (5, 7), (7, 5) \] 6. The minimum possible value of $a + b$ for a given $d$ is: \[ d(4 + 11) = 15d, \quad d(5 + 7) = 12d \] Clearly, this will be minimized when $d = 1$, so the answer is: \[ 12 \cdot 1 = 12 \] The final answer is $\boxed{12}$	12	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
[i]20 problems for 25 minutes.[/i] [b]p1.[/b] Matt has a twenty dollar bill and buys two items worth $\$7:99$ each. How much change does he receive, in dollars? [b]p2.[/b] The sum of two distinct numbers is equal to the positive difference of the two numbers. What is the product of the two numbers? [b]p3.[/b] Evaluate $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7}{8 + 9 + 10 + 11 + 12 + 13 + 14}.$$ [b]p4.[/b] A sphere with radius $r$ has volume $2\pi$. Find the volume of a sphere with diameter $r$. [b]p5.[/b] Yannick ran $100$ meters in $14.22$ seconds. Compute his average speed in meters per second, rounded to the nearest integer. [b]p6.[/b] The mean of the numbers $2, 0, 1, 5,$ and $x$ is an integer. Find the smallest possible positive integer value for $x$. [b]p7.[/b] Let $f(x) =\sqrt{2^2 - x^2}$. Find the value of $f(f(f(f(f(-1)))))$. [b]p8.[/b] Find the smallest positive integer $n$ such that $20$ divides $15n$ and $15$ divides $20n$. [b]p9.[/b] A circle is inscribed in equilateral triangle $ABC$. Let $M$ be the point where the circle touches side $AB$ and let $N$ be the second intersection of segment $CM$ and the circle. Compute the ratio $\frac{MN}{CN}$ . [b]p10.[/b] Four boys and four girls line up in a random order. What is the probability that both the first and last person in line is a girl? [b]p11.[/b] Let $k$ be a positive integer. After making $k$ consecutive shots successfully, Andy's overall shooting accuracy increased from $65\%$ to $70\%$. Determine the minimum possible value of $k$. [b]p12.[/b] In square $ABCD$, $M$ is the midpoint of side $CD$. Points $N$ and $P$ are on segments $BC$ and $AB$ respectively such that $ \angle AMN = \angle MNP = 90^o$. Compute the ratio $\frac{AP}{PB}$ . [b]p13.[/b] Meena writes the numbers $1, 2, 3$, and $4$ in some order on a blackboard, such that she cannot swap two numbers and obtain the sequence $1$, $2$, $3$, $4$. How many sequences could she have written? [b]p14.[/b] Find the smallest positive integer $N$ such that $2N$ is a perfect square and $3N$ is a perfect cube. [b]p15.[/b] A polyhedron has $60$ vertices, $150$ edges, and $92$ faces. If all of the faces are either regular pentagons or equilateral triangles, how many of the $92$ faces are pentagons? [b]p16.[/b] All positive integers relatively prime to $2015$ are written in increasing order. Let the twentieth number be $p$. The value of $\frac{2015}{p}-1$ can be expressed as $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Compute $a + b$. [b]p17.[/b] Five red lines and three blue lines are drawn on a plane. Given that $x$ pairs of lines of the same color intersect and $y$ pairs of lines of different colors intersect, find the maximum possible value of $y - x$. [b]p18.[/b] In triangle $ABC$, where $AC > AB$, $M$ is the midpoint of $BC$ and $D$ is on segment $AC$ such that $DM$ is perpendicular to $BC$. Given that the areas of $MAD$ and $MBD$ are $5$ and $6$, respectively, compute the area of triangle $ABC$. [b]p19.[/b] For how many ordered pairs $(x, y)$ of integers satisfying $0 \le x, y \le 10$ is $(x + y)^2 + (xy - 1)^2$ a prime number? [b]p20.[/b] A solitaire game is played with $8$ red, $9$ green, and $10$ blue cards. Totoro plays each of the cards exactly once in some order, one at a time. When he plays a card of color $c$, he gains a number of points equal to the number of cards that are not of color $c$ in his hand. Find the maximum number of points that he can obtain by the end of the game. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To solve the problem, we need to determine the number of ordered pairs $(x, y)$ of integers satisfying $0 \le x, y \le 10$ such that $(x + y)^2 + (xy - 1)^2$ is a prime number. 1. Start by expanding the given expression: \[ (x + y)^2 + (xy - 1)^2 = x^2 + 2xy + y^2 + x^2y^2 - 2xy + 1 = x^2 + y^2 + x^2y^2 + 1 \] Simplify the expression: \[ (x + y)^2 + (xy - 1)^2 = (x^2 + 1)(y^2 + 1) \] 2. For the product $(x^2 + 1)(y^2 + 1)$ to be a prime number, one of the factors must be 1 because a prime number has exactly two distinct positive divisors: 1 and itself. 3. Therefore, we need either $x^2 + 1 = 1$ or $y^2 + 1 = 1$: - If $x^2 + 1 = 1$, then $x^2 = 0$ which implies $x = 0$. - If $y^2 + 1 = 1$, then $y^2 = 0$ which implies $y = 0$. 4. Consider the case $x = 0$: - Then $y^2 + 1$ must be a prime number. - The possible values of $y$ within the range $0 \le y \le 10$ are those for which $y^2 + 1$ is prime: \[ y = 1 \implies y^2 + 1 = 2 \quad (\text{prime}) \] \[ y = 2 \implies y^2 + 1 = 5 \quad (\text{prime}) \] \[ y = 3 \implies y^2 + 1 = 10 \quad (\text{not prime}) \] \[ y = 4 \implies y^2 + 1 = 17 \quad (\text{prime}) \] \[ y = 5 \implies y^2 + 1 = 26 \quad (\text{not prime}) \] \[ y = 6 \implies y^2 + 1 = 37 \quad (\text{prime}) \] \[ y = 7 \implies y^2 + 1 = 50 \quad (\text{not prime}) \] \[ y = 8 \implies y^2 + 1 = 65 \quad (\text{not prime}) \] \[ y = 9 \implies y^2 + 1 = 82 \quad (\text{not prime}) \] \[ y = 10 \implies y^2 + 1 = 101 \quad (\text{prime}) \] - Thus, the valid values of $y$ are $1, 2, 4, 6, 10$, giving 5 ordered pairs $(0, 1), (0, 2), (0, 4), (0, 6), (0, 10)$. 5. Consider the case $y = 0$: - Then $x^2 + 1$ must be a prime number. - The possible values of $x$ within the range $0 \le x \le 10$ are those for which $x^2 + 1$ is prime: \[ x = 1 \implies x^2 + 1 = 2 \quad (\text{prime}) \] \[ x = 2 \implies x^2 + 1 = 5 \quad (\text{prime}) \] \[ x = 3 \implies x^2 + 1 = 10 \quad (\text{not prime}) \] \[ x = 4 \implies x^2 + 1 = 17 \quad (\text{prime}) \] \[ x = 5 \implies x^2 + 1 = 26 \quad (\text{not prime}) \] \[ x = 6 \implies x^2 + 1 = 37 \quad (\text{prime}) \] \[ x = 7 \implies x^2 + 1 = 50 \quad (\text{not prime}) \] \[ x = 8 \implies x^2 + 1 = 65 \quad (\text{not prime}) \] \[ x = 9 \implies x^2 + 1 = 82 \quad (\text{not prime}) \] \[ x = 10 \implies x^2 + 1 = 101 \quad (\text{prime}) \] - Thus, the valid values of $x$ are $1, 2, 4, 6, 10$, giving 5 ordered pairs $(1, 0), (2, 0), (4, 0), (6, 0), (10, 0)$. 6. Combining both cases, we have a total of $5 + 5 = 10$ ordered pairs. The final answer is $\boxed{10}$.	10	Other	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] A right triangle has a hypotenuse of length $25$ and a leg of length $16$. Compute the length of the other leg of this triangle. [b]p2.[/b] Tanya has a circular necklace with $5$ evenly-spaced beads, each colored red or blue. Find the number of distinct necklaces in which no two red beads are adjacent. If a necklace can be transformed into another necklace through a series of rotations and reflections, then the two necklaces are considered to be the same. [b]p3.[/b] Find the sum of the digits in the decimal representation of $10^{2016} - 2016$. [b]p4.[/b] Let $x$ be a real number satisfying $$x^1 \cdot x^2 \cdot x^3 \cdot x^4 \cdot x^5 \cdot x^6 = 8^7.$$ Compute the value of $x^7$. [b]p5.[/b] What is the smallest possible perimeter of an acute, scalene triangle with integer side lengths? [b]p6.[/b] Call a sequence $a_1, a_2, a_3,..., a_n$ mountainous if there exists an index $t$ between $1$ and $n$ inclusive such that $$a_1 \le a_2\le ... \le a_t \,\,\,\, and \,\,\,\, a_t \ge a_{t+1} \ge ... \ge a_n$$ In how many ways can Bishal arrange the ten numbers $1$, $1$, $2$, $2$, $3$, $3$, $4$, $4$, $5$, and $5$ into a mountainous sequence? (Two possible mountainous sequences are $1$, $1$, $2$, $3$, $4$, $4$, $5$, $5$, $3$, $2$ and $5$, $5$, $4$, $4$, $3$, $3$, $2$, $2$, $1$, $1$.) [b]p7.[/b] Find the sum of the areas of all (non self-intersecting) quadrilaterals whose vertices are the four points $(-3,-6)$, $(7,-1)$, $(-2, 9)$, and $(0, 0)$. [b]p8.[/b] Mohammed Zhang's favorite function is $f(x) =\sqrt{x^2 - 4x + 5} +\sqrt{x^2 + 4x + 8}$. Find the minumum possible value of $f(x)$ over all real numbers $x$. [b]p9.[/b] A segment $AB$ with length $1$ lies on a plane. Find the area of the set of points $P$ in the plane for which $\angle APB$ is the second smallest angle in triangle $ABP$. [b]p10.[/b] A binary string is a dipalindrome if it can be produced by writing two non-empty palindromic strings one after the other. For example, $10100100$ is a dipalindrome because both $101$ and $00100$ are palindromes. How many binary strings of length $18$ are both palindromes and dipalindromes? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To find the minimum value of the function $ f(x) = \sqrt{x^2 - 4x + 5} + \sqrt{x^2 + 4x + 8} $, we can interpret the terms geometrically. 1. Rewrite the terms inside the square roots: \[ \sqrt{x^2 - 4x + 5} = \sqrt{(x-2)^2 + 1^2} \] \[ \sqrt{x^2 + 4x + 8} = \sqrt{(x+2)^2 + 2^2} \] This allows us to interpret the terms as distances in the coordinate plane. 2. Define points based on these distances: \[ A = (x, 0), \quad B = (2, 1), \quad C = (-2, 2) \] The expression $ f(x) $ represents the sum of the distances from point $ A $ to points $ B $ and $ C $. 3. Reflect point $ B $ over the $ x $-axis to get $ B' $: \[ B' = (2, -1) \] The distance $ AB' $ is the same as $ AB $ because the reflection does not change the distance. 4. The sum of the distances $ AB' + AC $ will be minimized when point $ A $ is collinear with points $ B' $ and $ C $. 5. Find the equation of the line passing through $ B' $ and $ C $: \[ \text{slope of } B'C = \frac{2 - (-1)}{-2 - 2} = \frac{3}{-4} = -\frac{3}{4} \] The equation of the line in point-slope form is: \[ y - (-1) = -\frac{3}{4}(x - 2) \] Simplifying, we get: \[ y + 1 = -\frac{3}{4}x + \frac{3}{2} \] \[ y = -\frac{3}{4}x + \frac{1}{2} \] 6. Find the $ x $-intercept of this line (where $ y = 0 $): \[ 0 = -\frac{3}{4}x + \frac{1}{2} \] \[ \frac{3}{4}x = \frac{1}{2} \] \[ x = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3} \] 7. Evaluate $ f(x) $ at $ x = \frac{2}{3} $: \[ f\left(\frac{2}{3}\right) = \sqrt{\left(\frac{2}{3} - 2\right)^2 + 1^2} + \sqrt{\left(\frac{2}{3} + 2\right)^2 + 2^2} \] \[ = \sqrt{\left(-\frac{4}{3}\right)^2 + 1} + \sqrt{\left(\frac{8}{3}\right)^2 + 4} \] \[ = \sqrt{\frac{16}{9} + 1} + \sqrt{\frac{64}{9} + 4} \] \[ = \sqrt{\frac{16}{9} + \frac{9}{9}} + \sqrt{\frac{64}{9} + \frac{36}{9}} \] \[ = \sqrt{\frac{25}{9}} + \sqrt{\frac{100}{9}} \] \[ = \frac{5}{3} + \frac{10}{3} \] \[ = \frac{15}{3} = 5 \] Thus, the minimum value of $ f(x) $ is $ \boxed{5} $.	5	Geometry	math-word-problem	Yes	Yes	aops_forum	false
[i]20 problems for 25 minutes.[/i] [b]p1.[/b] Compute the value of $2 + 20 + 201 + 2016$. [b]p2.[/b] Gleb is making a doll, whose prototype is a cube with side length $5$ centimeters. If the density of the toy is $4$ grams per cubic centimeter, compute its mass in grams. [b]p3.[/b] Find the sum of $20\%$ of $16$ and $16\%$ of $20$. [b]p4.[/b] How many times does Akmal need to roll a standard six-sided die in order to guarantee that two of the rolled values sum to an even number? [b]p5.[/b] During a period of one month, there are ten days without rain and twenty days without snow. What is the positive difference between the number of rainy days and the number of snowy days? [b]p6.[/b] Joanna has a fully charged phone. After using it for $30$ minutes, she notices that $20$ percent of the battery has been consumed. Assuming a constant battery consumption rate, for how many additional minutes can she use the phone until $20$ percent of the battery remains? [b]p7.[/b] In a square $ABCD$, points $P$, $Q$, $R$, and $S$ are chosen on sides $AB$, $BC$, $CD$, and $DA$ respectively, such that $AP = 2PB$, $BQ = 2QC$, $CR = 2RD$, and $DS = 2SA$. What fraction of square $ABCD$ is contained within square $PQRS$? [b]p8.[/b] The sum of the reciprocals of two not necessarily distinct positive integers is $1$. Compute the sum of these two positive integers. [b]p9.[/b] In a room of government officials, two-thirds of the men are standing and $8$ women are standing. There are twice as many standing men as standing women and twice as many women in total as men in total. Find the total number of government ocials in the room. [b]p10.[/b] A string of lowercase English letters is called pseudo-Japanese if it begins with a consonant and alternates between consonants and vowels. (Here the letter "y" is considered neither a consonant nor vowel.) How many $4$-letter pseudo-Japanese strings are there? [b]p11.[/b] In a wooden box, there are $2$ identical black balls, $2$ identical grey balls, and $1$ white ball. Yuka randomly draws two balls in succession without replacement. What is the probability that the first ball is strictly darker than the second one? [b]p12.[/b] Compute the real number $x$ for which $(x + 1)^2 + (x + 2)^2 + (x + 3)^2 = (x + 4)^2 + (x + 5)^2 + (x + 6)^2$. [b]p13.[/b] Let $ABC$ be an isosceles right triangle with $\angle C = 90^o$ and $AB = 2$. Let $D$, $E$, and $F$ be points outside $ABC$ in the same plane such that the triangles $DBC$, $AEC$, and $ABF$ are isosceles right triangles with hypotenuses $BC$, $AC$, and $AB$, respectively. Find the area of triangle $DEF$. [b]p14.[/b] Salma is thinking of a six-digit positive integer $n$ divisible by $90$. If the sum of the digits of n is divisible by $5$, find $n$. [b]p15.[/b] Kiady ate a total of $100$ bananas over five days. On the ($i + 1$)-th day ($1 \le i \le 4$), he ate i more bananas than he did on the $i$-th day. How many bananas did he eat on the fifth day? [b]p16.[/b] In a unit equilateral triangle $ABC$; points $D$,$E$, and $F$ are chosen on sides $BC$, $CA$, and $AB$, respectively. If lines $DE$, $EF$, and $FD$ are perpendicular to $CA$, $AB$ and $BC$, respectively, compute the area of triangle $DEF$. [b]p17.[/b] Carlos rolls three standard six-sided dice. What is the probability that the product of the three numbers on the top faces has units digit 5? [b]p18.[/b] Find the positive integer $n$ for which $n^{n^n}= 3^{3^{82}}$. [b]p19.[/b] John folds a rope in half five times then cuts the folded rope with four knife cuts, leaving five stacks of rope segments. How many pieces of rope does he now have? [b]p20.[/b] An integer $n > 1$ is conglomerate if all positive integers less than n and relatively prime to $n$ are not composite. For example, $3$ is conglomerate since $1$ and $2$ are not composite. Find the sum of all conglomerate integers less than or equal to $200$. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To solve the problem, we need to find the area of triangle $ DEF $ given the specific geometric conditions. Let's break down the problem step by step. 1. Understanding the Geometry: - We are given an isosceles right triangle $ \triangle ABC $ with $ \angle C = 90^\circ $ and $ AB = 2 $. - Points $ D, E, $ and $ F $ are such that $ \triangle DBC, \triangle AEC, $ and $ \triangle ABF $ are isosceles right triangles with hypotenuses $ BC, AC, $ and $ AB $ respectively. 2. Determine the Side Lengths: - Since $ \triangle ABC $ is an isosceles right triangle with $ AB = 2 $, the legs $ AC $ and $ BC $ are equal. Using the Pythagorean theorem: \[ AC = BC = \frac{AB}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \] 3. Constructing the Isosceles Right Triangles: - $ \triangle DBC $ is an isosceles right triangle with hypotenuse $ BC = \sqrt{2} $. Therefore, the legs $ BD $ and $ DC $ are: \[ BD = DC = \frac{BC}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1 \] - Similarly, $ \triangle AEC $ and $ \triangle ABF $ are isosceles right triangles with hypotenuses $ AC $ and $ AB $ respectively. Thus: \[ AE = EC = 1 \quad \text{and} \quad AF = BF = \frac{AB}{\sqrt{2}} = \sqrt{2} \] 4. Forming the Rectangle: - Since $ \triangle ABC $ is isosceles and right, and the triangles $ \triangle DBC, \triangle AEC, $ and $ \triangle ABF $ are isosceles right triangles, the points $ D, E, $ and $ F $ form a rectangle $ AFBDE $ with: \[ AB = 2 \quad \text{and} \quad BD = 1 \] 5. Calculating the Area of $ \triangle DEF $: - The triangle $ DEF $ is formed within the rectangle $ AFBDE $. The base $ EF $ of $ \triangle DEF $ is equal to $ AB = 2 $, and the height from $ D $ to $ EF $ is also $ 2 $ (since $ D $ is directly above the midpoint of $ EF $). - The area of $ \triangle DEF $ is given by: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2 \] $\blacksquare$ The final answer is $ \boxed{2} $.	2	Other	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] On SeaBay, green herring costs $\$2.50$ per pound, blue herring costs $\$4.00$ per pound, and red herring costs $\$5,85$ per pound. What must Farmer James pay for $12$ pounds of green herring and $7$ pounds of blue herring, in dollars? [b]p2.[/b] A triangle has side lengths $3$, $4$, and $6$. A second triangle, similar to the first one, has one side of length $12$. Find the sum of all possible lengths of the second triangle's longest side. [b]p3.[/b] Hen Hao runs two laps around a track. Her overall average speed for the two laps was $20\%$ slower than her average speed for just the first lap. What is the ratio of Hen Hao's average speed in the first lap to her average speed in the second lap? [b]p4.[/b] Square $ABCD$ has side length $2$. Circle $\omega$ is centered at $A$ with radius $2$, and intersects line $AD$ at distinct points $D$ and $E$. Let $X$ be the intersection of segments $EC$ and $AB$, and let $Y$ be the intersection of the minor arc $DB$ with segment $EC$. Compute the length of $XY$ . [b]p5.[/b] Hen Hao rolls $4$ tetrahedral dice with faces labeled $1$, $2$, $3$, and $4$, and adds up the numbers on the faces facing down. Find the probability that she ends up with a sum that is a perfect square. [b]p6.[/b] Let $N \ge 11$ be a positive integer. In the Eggs-Eater Lottery, Farmer James needs to choose an (unordered) group of six different integers from $1$ to $N$, inclusive. Later, during the live drawing, another group of six numbers from $1$ to $N$ will be randomly chosen as winning numbers. Farmer James notices that the probability he will choose exactly zero winning numbers is the same as the probability that he will choose exactly one winning number. What must be the value of $N$? [b]p7.[/b] An egg plant is a hollow cylinder of negligible thickness with radius $2$ and height $h$. Inside the egg plant, there is enough space for four solid spherical eggs of radius $1$. What is the minimum possible value for $h$? [b]p8.[/b] Let $a_1, a_2, a_3, ...$ be a geometric sequence of positive reals such that $a_1 < 1$ and $(a_{20})^{20} = (a_{18})^{18}$. What is the smallest positive integer n such that the product $a_1a_2a_3...a_n$ is greater than $1$? [b]p9.[/b] In parallelogram $ABCD$, the angle bisector of $\angle DAB$ meets segment $BC$ at $E$, and $AE$ and $BD$ intersect at $P$. Given that $AB = 9$, $AE = 16$, and $EP = EC$, find $BC$. [b]p10.[/b] Farmer James places the numbers $1, 2,..., 9$ in a $3\times 3$ grid such that each number appears exactly once in the grid. Let $x_i$ be the product of the numbers in row $i$, and $y_i$ be the product of the numbers in column $i$. Given that the unordered sets $\{x_1, x_2, x_3\}$ and $\{y_1, y_2, y_3\}$ are the same, how many possible arrangements could Farmer James have made? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	### Problem 1 1. Calculate the cost for 12 pounds of green herring: \[ 12 \text{ pounds} \times \$2.50 \text{ per pound} = 12 \times 2.50 = \$30.00 \] 2. Calculate the cost for 7 pounds of blue herring: \[ 7 \text{ pounds} \times \$4.00 \text{ per pound} = 7 \times 4.00 = \$28.00 \] 3. Add the costs together to find the total amount Farmer James must pay: \[ \$30.00 + \$28.00 = \$58.00 \] The final answer is $\boxed{0}$	0	Other	math-word-problem	Yes	Yes	aops_forum	false
[u]Round 1[/u] [b]p1.[/b] What is the remainder when $2021$ is divided by $102$? [b]p2.[/b] Brian has $2$ left shoes and $2$ right shoes. Given that he randomly picks $2$ of the $4$ shoes, the probability he will get a left shoe and a right shoe is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Compute $p + q$. [b]p3.[/b] In how many ways can $59$ be written as a sum of two perfect squares? (The order of the two perfect squares does not matter.) [u]Round 2 [/u] [b]p4.[/b] Two positive integers have a sum of $60$. Their least common multiple is $273$. What is the positive diffeerence between the two numbers? [b]p5.[/b] How many ways are there to distribute $13$ identical apples among $4$ identical boxes so that no two boxes receive the same number of apples? A box may receive zero apples. [b]p6.[/b] In square $ABCD$ with side length $5$, $P$ lies on segment $AB$ so that $AP = 3$ and $Q$ lies on segment $AD$ so that $AQ = 4$. Given that the area of triangle $CPQ$ is $x$, compute $2x$. [u]Round 3 [/u] [b]p7.[/b] Find the number of ordered triples $(a, b, c)$ of nonnegative integers such that $2a+3b+5c = 15$. [b]p8.[/b] What is the greatest integer $n \le 15$ such that $n + 1$ and $n^2 + 3$ are both prime? [b]p9.[/b] For positive integers $a, b$, and $c$, suppose that $gcd \,\,(a, b) = 21$, $gcd \,\,(a, c) = 10$, and $gcd \,\,(b,c) = 11$. Find $\frac{abc}{lcm \,\,(a,b,c)}$ . (Note: $gcd$ is the greatest common divisor function and $lcm$ is the least common multiple function.) [u]Round 4[/u] [b]p10.[/b] The vertices of a square in the coordinate plane are at $(0, 0)$, $(0, 6)$, $(6, 0)$, and $(6, 6)$. Line $\ell$ intersects the square at exactly two lattice points (that is, points with integer coordinates). How many such lines $\ell$ are there that divide the square into two regions, one of them having an area of $12$? [b]p11.[/b] Let $f(n)$ be defined as follows for positive integers $n$: $f(1) = 0$, $f(n) = 1$ if $n$ is prime, and $f(n) = f(n - 1) + 1$ otherwise. What is the maximum value of $f(n)$ for $n \le 120$? [b]p12.[/b] The graph of the equation $y = x^3 + ax^2 + bx + c$ passes through the points $(2,4)$, $(3, 9)$, and $(4, 16)$. What is $b$? PS. You should use hide for answers. Rounds 5- 8 have been posted [url=https://artofproblemsolving.com/community/c3h2949415p26408227]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	### Problem 1 1. We need to find the remainder when $2021$ is divided by $102$. 2. Perform the division: \[ 2021 \div 102 \approx 19.814 \] 3. The integer part of the division is $19$, so: \[ 2021 = 102 \times 19 + r \] 4. Calculate the remainder $r$: \[ r = 2021 - 102 \times 19 = 2021 - 1938 = 83 \] The final answer is $\boxed{8}$.	8	Other	math-word-problem	Yes	Yes	aops_forum	false
Two positive whole numbers differ by $3$. The sum of their squares is $117$. Find the larger of the two numbers.	1. Let the two positive whole numbers be $ a $ and $ b $. Given that they differ by 3, we can write: \[ a - b = 3 \] Without loss of generality, assume $ a > b $. Therefore, we can write: \[ a = b + 3 \] 2. We are also given that the sum of their squares is 117: \[ a^2 + b^2 = 117 \] 3. Substitute $ a = b + 3 $ into the equation $ a^2 + b^2 = 117 $: \[ (b + 3)^2 + b^2 = 117 \] 4. Expand and simplify the equation: \[ (b + 3)^2 + b^2 = 117 \] \[ b^2 + 6b + 9 + b^2 = 117 \] \[ 2b^2 + 6b + 9 = 117 \] 5. Subtract 117 from both sides to set the equation to zero: \[ 2b^2 + 6b + 9 - 117 = 0 \] \[ 2b^2 + 6b - 108 = 0 \] 6. Divide the entire equation by 2 to simplify: \[ b^2 + 3b - 54 = 0 \] 7. Solve the quadratic equation using the quadratic formula $ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} $, where $ A = 1 $, $ B = 3 $, and $ C = -54 $: \[ b = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-54)}}{2 \cdot 1} \] \[ b = \frac{-3 \pm \sqrt{9 + 216}}{2} \] \[ b = \frac{-3 \pm \sqrt{225}}{2} \] \[ b = \frac{-3 \pm 15}{2} \] 8. This gives us two potential solutions for $ b $: \[ b = \frac{-3 + 15}{2} = \frac{12}{2} = 6 \] \[ b = \frac{-3 - 15}{2} = \frac{-18}{2} = -9 \] 9. Since $ b $ must be a positive whole number, we discard $ b = -9 $ and accept $ b = 6 $. 10. Substitute $ b = 6 $ back into $ a = b + 3 $ to find $ a $: \[ a = 6 + 3 = 9 \] Conclusion: \[ \boxed{9} \]	9	Algebra	math-word-problem	Yes	Yes	aops_forum	false
The ages of Mr. and Mrs. Fibonacci are both two-digit numbers. If Mr. Fibonacci’s age can be formed by reversing the digits of Mrs. Fibonacci’s age, find the smallest possible positive difference between their ages.	1. Let Mr. Fibonacci's age be represented by the two-digit number $10a + b$, where $a$ and $b$ are the tens and units digits, respectively. 2. Let Mrs. Fibonacci's age be represented by the two-digit number $10b + a$, where $b$ and $a$ are the tens and units digits, respectively. 3. We need to find the smallest possible positive difference between their ages, which is given by: \[ \|(10a + b) - (10b + a)\| \] 4. Simplify the expression inside the absolute value: \[ \|(10a + b) - (10b + a)\| = \|10a + b - 10b - a\| = \|9a - 9b\| = 9\|a - b\| \] 5. To minimize the positive difference, we need to minimize $\|a - b\|$. Since $a$ and $b$ are digits (0 through 9), the smallest non-zero value for $\|a - b\|$ is 1. 6. Therefore, the smallest possible positive difference is: \[ 9 \times 1 = 9 \] 7. To verify, consider the ages where $\|a - b\| = 1$. For example, if $a = 4$ and $b = 5$, then Mr. Fibonacci's age is 45 and Mrs. Fibonacci's age is 54. The difference is: \[ \|54 - 45\| = 9 \] The final answer is $\boxed{9}$	9	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Tim is thinking of a positive integer between $2$ and $15,$ inclusive, and Ted is trying to guess the integer. Tim tells Ted how many factors his integer has, and Ted is then able to be certain of what Tim's integer is. What is Tim's integer?	1. We need to determine the number of factors for each integer between $2$ and $15$. 2. The number of factors of a number $n$ can be determined by its prime factorization. If $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, then the number of factors of $n$ is given by $(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$. Let's calculate the number of factors for each integer from $2$ to $15$: - $2 = 2^1$: $(1+1) = 2$ factors. - $3 = 3^1$: $(1+1) = 2$ factors. - $4 = 2^2$: $(2+1) = 3$ factors. - $5 = 5^1$: $(1+1) = 2$ factors. - $6 = 2 \times 3$: $(1+1)(1+1) = 2 \times 2 = 4$ factors. - $7 = 7^1$: $(1+1) = 2$ factors. - $8 = 2^3$: $(3+1) = 4$ factors. - $9 = 3^2$: $(2+1) = 3$ factors. - $10 = 2 \times 5$: $(1+1)(1+1) = 2 \times 2 = 4$ factors. - $11 = 11^1$: $(1+1) = 2$ factors. - $12 = 2^2 \times 3$: $(2+1)(1+1) = 3 \times 2 = 6$ factors. - $13 = 13^1$: $(1+1) = 2$ factors. - $14 = 2 \times 7$: $(1+1)(1+1) = 2 \times 2 = 4$ factors. - $15 = 3 \times 5$: $(1+1)(1+1) = 2 \times 2 = 4$ factors. 3. From the calculations, we see that the number $12$ is the only number between $2$ and $15$ that has exactly $6$ factors. Conclusion: Since Tim's integer is the only one with $6$ factors, Ted can be certain that Tim's integer is $12$. The final answer is $\boxed{12}$.	12	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Al travels for $20$ miles per hour rolling down a hill in his chair for two hours, then four miles per hour climbing a hill for six hours. What is his average speed, in miles per hour?	1. Calculate the total distance Al travels: - Rolling down the hill: \[ 20 \text{ miles per hour} \times 2 \text{ hours} = 40 \text{ miles} \] - Climbing the hill: \[ 4 \text{ miles per hour} \times 6 \text{ hours} = 24 \text{ miles} \] - Total distance: \[ D_{\text{total}} = 40 \text{ miles} + 24 \text{ miles} = 64 \text{ miles} \] 2. Calculate the total time Al spends traveling: - Rolling down the hill: \[ 2 \text{ hours} \] - Climbing the hill: \[ 6 \text{ hours} \] - Total time: \[ T_{\text{total}} = 2 \text{ hours} + 6 \text{ hours} = 8 \text{ hours} \] 3. Calculate the average speed: \[ \text{Average speed} = \frac{D_{\text{total}}}{T_{\text{total}}} = \frac{64 \text{ miles}}{8 \text{ hours}} = 8 \text{ miles per hour} \] The final answer is $\boxed{8}$ miles per hour.	8	Algebra	math-word-problem	Yes	Yes	aops_forum	false
Carl, James, Saif, and Ted play several games of two-player For The Win on the Art of Problem Solving website. If, among these games, Carl wins $5$ and loses $0,$ James wins $4$ and loses $2,$ Saif wins $1$ and loses $6,$ and Ted wins $4,$ how many games does Ted lose?	1. We start by noting that each game has exactly one winner and one loser. Therefore, the total number of wins must equal the total number of losses. 2. We are given the number of wins for each player: - Carl wins 5 games. - James wins 4 games. - Saif wins 1 game. - Ted wins 4 games. 3. Summing these, the total number of wins is: \[ 5 + 4 + 1 + 4 = 14 \] 4. We are also given the number of losses for some players: - Carl loses 0 games. - James loses 2 games. - Saif loses 6 games. - Ted's losses are unknown and need to be determined. 5. Let $ L $ represent the number of games Ted loses. Since the total number of losses must equal the total number of wins, we set up the equation: \[ 0 + 2 + 6 + L = 14 \] 6. Simplifying the left-hand side, we get: \[ 8 + L = 14 \] 7. Solving for $ L $, we subtract 8 from both sides: \[ L = 14 - 8 = 6 \] Thus, Ted loses 6 games. The final answer is $\boxed{6}$.	6	Logic and Puzzles	math-word-problem	Yes	Yes	aops_forum	false
Seongcheol has $3$ red shirts and $2$ green shirts, such that he cannot tell the difference between his three red shirts and he similarly cannot tell the difference between his two green shirts. In how many ways can he hang them in a row in his closet, given that he does not want the two green shirts next to each other?	1. Calculate the total number of ways to arrange the shirts without any restrictions: Seongcheol has 3 red shirts (R) and 2 green shirts (G). The total number of shirts is 5. Since the red shirts are indistinguishable among themselves and the green shirts are indistinguishable among themselves, the number of ways to arrange these shirts is given by the multinomial coefficient: \[ \frac{5!}{3!2!} = \frac{120}{6 \cdot 2} = 10 \] 2. Calculate the number of ways to arrange the shirts with the two green shirts next to each other: If the two green shirts are next to each other, we can treat them as a single "block" or "big green shirt". This reduces the problem to arranging 4 items: 3 red shirts and 1 "big green shirt". The number of ways to arrange these 4 items is: \[ \frac{4!}{3!1!} = \frac{24}{6} = 4 \] 3. Subtract the restricted arrangements from the total arrangements: The number of ways to arrange the shirts such that the two green shirts are not next to each other is the total number of arrangements minus the number of restricted arrangements: \[ 10 - 4 = 6 \] Conclusion: \[ \boxed{6} \]	6	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
Let $f(x)=x^2-2x+1.$ For some constant $k, f(x+k) = x^2+2x+1$ for all real numbers $x.$ Determine the value of $k.$	1. Given the function $ f(x) = x^2 - 2x + 1 $, we can rewrite it in a factored form: \[ f(x) = (x-1)^2 \] 2. We are given that for some constant $ k $, the function $ f(x+k) $ is equal to $ x^2 + 2x + 1 $ for all real numbers $ x $. Therefore, we need to find $ k $ such that: \[ f(x+k) = x^2 + 2x + 1 \] 3. Substitute $ x+k $ into the function $ f $: \[ f(x+k) = (x+k-1)^2 \] 4. Expand the expression: \[ (x+k-1)^2 = (x+k-1)(x+k-1) = x^2 + 2x(k-1) + (k-1)^2 \] 5. We need this to be equal to $ x^2 + 2x + 1 $: \[ x^2 + 2x(k-1) + (k-1)^2 = x^2 + 2x + 1 \] 6. By comparing the coefficients of $ x $ and the constant terms on both sides of the equation, we get: \[ 2x(k-1) = 2x \quad \text{and} \quad (k-1)^2 = 1 \] 7. From the coefficient of $ x $: \[ 2(k-1) = 2 \implies k-1 = 1 \implies k = 2 \] 8. From the constant term: \[ (k-1)^2 = 1 \implies k-1 = \pm 1 \] This gives us two possible solutions: \[ k-1 = 1 \implies k = 2 \] \[ k-1 = -1 \implies k = 0 \] 9. However, since $ k = 0 $ does not satisfy the coefficient comparison $ 2(k-1) = 2 $, the only valid solution is: \[ k = 2 \] The final answer is $ \boxed{2} $	2	Algebra	math-word-problem	Yes	Yes	aops_forum	false
$25.$ Let $C$ be the answer to Problem $27.$ What is the $C$-th smallest positive integer with exactly four positive factors? $26.$ Let $A$ be the answer to Problem $25.$ Determine the absolute value of the difference between the two positive integer roots of the quadratic equation $x^2-Ax+48=0$ $27.$ Let $B$ be the answer to Problem $26.$ Compute the smallest integer greater than $\frac{B}{\pi}$	To solve the given set of problems, we need to follow the sequence of dependencies between the problems. Let's break down each problem step-by-step. ### Problem 25: We need to find the $ C $-th smallest positive integer with exactly four positive factors. A positive integer $ n $ has exactly four positive factors if and only if $ n $ is of the form $ p^3 $ or $ p \cdot q $ where $ p $ and $ q $ are distinct prime numbers. Let's list the smallest integers with exactly four positive factors: 1. $ 6 = 2 \times 3 $ 2. $ 8 = 2^3 $ 3. $ 10 = 2 \times 5 $ 4. $ 14 = 2 \times 7 $ 5. $ 15 = 3 \times 5 $ 6. $ 21 = 3 \times 7 $ 7. $ 22 = 2 \times 11 $ 8. $ 26 = 2 \times 13 $ Thus, the 8th smallest positive integer with exactly four positive factors is $ 26 $. ### Problem 26: Let $ A $ be the answer to Problem 25, which is $ 26 $. We need to determine the absolute value of the difference between the two positive integer roots of the quadratic equation $ x^2 - Ax + 48 = 0 $. The quadratic equation is: \[ x^2 - 26x + 48 = 0 \] To find the roots, we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where $ a = 1 $, $ b = -26 $, and $ c = 48 $. Substituting the values, we get: \[ x = \frac{26 \pm \sqrt{26^2 - 4 \cdot 1 \cdot 48}}{2 \cdot 1} \] \[ x = \frac{26 \pm \sqrt{676 - 192}}{2} \] \[ x = \frac{26 \pm \sqrt{484}}{2} \] \[ x = \frac{26 \pm 22}{2} \] Thus, the roots are: \[ x = \frac{26 + 22}{2} = 24 \] \[ x = \frac{26 - 22}{2} = 2 \] The absolute value of the difference between the roots is: \[ \|24 - 2\| = 22 \] ### Problem 27: Let $ B $ be the answer to Problem 26, which is $ 22 $. We need to compute the smallest integer greater than $ \frac{B}{\pi} $. \[ \frac{22}{\pi} \approx \frac{22}{3.14159} \approx 7.0028 \] The smallest integer greater than $ 7.0028 $ is $ 8 $. The final answer is $ \boxed{ 8 } $.	8	Logic and Puzzles	math-word-problem	Yes	Yes	aops_forum	false
[u]Round 1[/u] [b]p1.[/b] How many powers of $2$ are greater than $3$ but less than $2013$? [b]p2.[/b] What number is equal to six greater than three times the answer to this question? [b]p3.[/b] Surya Cup-a-tea-raju goes to Starbucks Coffee to sip coffee out of a styrofoam cup. The cup is a cylinder, open on one end, with base radius $3$ centimeters and height $10$ centimeters. What is the exterior surface area of the styrofoam cup? [u]Round 2[/u] [b]p4.[/b] Andrew has two $6$-foot-length sticks that he wishes to make into two of the sides of the entrance to his fort, with the ground being the third side. If he wants to make his entrance in the shape of a triangle, what is the largest area that he can make the entrance? [b]p5.[/b] Ethan and Devin met a fairy who told them “if you have less than $15$ dollars, I will give you cake”. If both had integral amounts of dollars, and Devin had 5 more dollars than Ethan, but only Ethan got cake, how many different amounts of money could Ethan have had? [b]p6.[/b] If $2012^x = 2013$, for what value of $a$, in terms of $x$, is it true that $2012^a = 2013^2$? [u]Round 3[/u] [b]p7.[/b] Find the ordered triple $(L, M, T)$ of positive integers that makes the following equation true: $$1 + \dfrac{1}{L + \dfrac{1}{M+\dfrac{1}{T}}}=\frac{79}{43}.$$ [b]p8.[/b] Jonathan would like to start a banana plantation so he is saving up to buy an acre of land, which costs $\$600,000$. He deposits $\$300,000$ in the bank, which gives $20\%$ interest compounded at the end of each year. At this rate, how many years will Jonathan have to wait until he can buy the acre of land? [b]p9.[/b] Arul and Ethan went swimming at their town pool and started to swim laps to see who was in better shape. After one hour of swimming at their own paces, Ethan completed $32$ more laps than Arul. However, after that, Ethan got tired and swam at half his original speed while Arul’s speed didn’t change. After one more hour, Arul swam a total of $320$ laps. How many laps did Ethan swim after two hours? [u]Round 4[/u] [b]p10.[/b] A right triangle with a side length of $6$ and a hypotenuse of 10 has circles of radius $1$ centered at each vertex. What is the area of the space inside the triangle but outside all three circles? [b]p11.[/b] In isosceles trapezoid $ABCD$, $\overline{AB} \parallel\overline{CD}$ and the lengths of $\overline{AB}$ and $\overline{CD}$ are $2$ and $6$, respectively. Let the diagonals of the trapezoid intersect at point $E$. If the distance from $E$ to $\overline{CD}$ is $9$, what is the area of triangle $ABE$? [b]p12.[/b] If $144$ unit cubes are glued together to form a rectangular prism and the perimeter of the base is $54$ units, what is the height? PS. You should use hide for answers. Rounds 6-8 are [url=https://artofproblemsolving.com/community/c3h3136014p28427163]here[/url] and 9-12 [url=https://artofproblemsolving.com/community/c3h3137069p28442224]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. Finding the powers of $2$ greater than $3$ but less than $2013$ We need to find the integer values of $n$ such that: \[ 2^n > 3 \quad \text{and} \quad 2^n < 2013 \] First, solve $2^n > 3$: \[ n > \log_2 3 \approx 1.58496 \] So, $n \geq 2$. Next, solve $2^n < 2013$: \[ n < \log_2 2013 \approx 10.973 \] So, $n \leq 10$. Therefore, $n$ ranges from $2$ to $10$, inclusive. The number of integers in this range is: \[ 10 - 2 + 1 = 9 \] Conclusion: \[ \boxed{9} \] 2. Finding the number equal to six greater than three times the answer to the previous question Let $x$ be the answer to the previous question, which is $9$. We need to find: \[ 3x + 6 \] Substituting $x = 9$: \[ 3 \cdot 9 + 6 = 27 + 6 = 33 \] Conclusion: \[ \boxed{33} \] 3. Finding the exterior surface area of the styrofoam cup The cup is a cylinder with base radius $r = 3$ cm and height $h = 10$ cm. The exterior surface area of an open cylinder is given by: \[ 2\pi rh + \pi r^2 \] Substituting the given values: \[ 2\pi \cdot 3 \cdot 10 + \pi \cdot 3^2 = 60\pi + 9\pi = 69\pi \] Conclusion: \[ \boxed{69\pi} \] 4. Maximizing the area of a triangle with sides $6$ feet each and the ground as the third side The area of a triangle is maximized when it is a right triangle. For a right triangle with legs $a$ and $b$, the area is: \[ \frac{1}{2}ab \] Given $a = b = 6$ feet, the area is: \[ \frac{1}{2} \cdot 6 \cdot 6 = 18 \] Conclusion: \[ \boxed{18} \] 5. Finding the possible amounts of money Ethan could have had Let $E$ be the amount of money Ethan had, and $D$ be the amount Devin had. Given: \[ D = E + 5 \] Ethan got cake, so $E < 15$. Devin did not get cake, so $D \geq 15$. Therefore: \[ E + 5 \geq 15 \implies E \geq 10 \] Combining the inequalities: \[ 10 \leq E < 15 \] The possible values for $E$ are $10, 11, 12, 13, 14$. Conclusion: \[ \boxed{5} \] 6. Finding the value of $a$ such that $2012^a = 2013^2$ Given $2012^x = 2013$, we need to find $a$ such that: \[ 2012^a = 2013^2 \] Using the given equation: \[ 2013 = 2012^x \implies 2013^2 = (2012^x)^2 = 2012^{2x} \] Therefore: \[ a = 2x \] Conclusion: \[ \boxed{2x} \] 7. Finding the ordered triple $(L, M, T)$ that satisfies the equation Given: \[ 1 + \frac{1}{L + \frac{1}{M + \frac{1}{T}}} = \frac{79}{43} \] Subtract $1$ from both sides: \[ \frac{1}{L + \frac{1}{M + \frac{1}{T}}} = \frac{79}{43} - 1 = \frac{79 - 43}{43} = \frac{36}{43} \] Taking the reciprocal: \[ L + \frac{1}{M + \frac{1}{T}} = \frac{43}{36} \] Therefore, $L = 1$. Now: \[ \frac{1}{M + \frac{1}{T}} = \frac{43}{36} - 1 = \frac{43 - 36}{36} = \frac{7}{36} \] Taking the reciprocal: \[ M + \frac{1}{T} = \frac{36}{7} \] Therefore, $M = 5$. Now: \[ \frac{1}{T} = \frac{36}{7} - 5 = \frac{36 - 35}{7} = \frac{1}{7} \] Therefore, $T = 7$. Conclusion: \[ \boxed{(1, 5, 7)} \] 8. Finding the number of years until Jonathan can buy the acre of land Jonathan deposits $300,000 at $20\%$ interest compounded annually. We need to find the smallest $n$ such that: \[ 300,000 \times (1.2)^n \geq 600,000 \] Dividing both sides by 300,000: \[ (1.2)^n \geq 2 \] Taking the logarithm of both sides: \[ n \log(1.2) \geq \log(2) \] Solving for $n$: \[ n \geq \frac{\log(2)}{\log(1.2)} \approx \frac{0.3010}{0.07918} \approx 3.8 \] Therefore, $n = 4$. Conclusion: \[ \boxed{4} \] 9. Finding the number of laps Ethan swam after two hours Let $A$ be the number of laps Arul swam in one hour. Given: \[ A = 160 \] Ethan swam $32$ more laps than Arul in the first hour: \[ E_1 = A + 32 = 160 + 32 = 192 \] In the second hour, Ethan swam at half his original speed: \[ E_2 = \frac{192}{2} = 96 \] Total laps Ethan swam: \[ E = E_1 + E_2 = 192 + 96 = 288 \] Conclusion: \[ \boxed{288} \] 10. Finding the area inside the triangle but outside all three circles The right triangle has side lengths $6$ and $8$ (since $6^2 + 8^2 = 10^2$). The area of the triangle is: \[ \frac{1}{2} \cdot 6 \cdot 8 = 24 \] Each circle has radius $1$. The area of each circle is: \[ \pi \cdot 1^2 = \pi \] The total area of the three circles is: \[ 3\pi \] The area of the sectors inside the triangle is half the area of one circle: \[ \frac{3\pi}{2} \] The area inside the triangle but outside the circles is: \[ 24 - \frac{3\pi}{2} \] Conclusion: \[ \boxed{24 - \frac{3\pi}{2}} \] 11. Finding the area of triangle $ABE$ Given isosceles trapezoid $ABCD$ with $\overline{AB} \parallel \overline{CD}$, $\overline{AB} = 2$, $\overline{CD} = 6$, and the distance from $E$ to $\overline{CD}$ is $9$. By similar triangles, the altitude from $E$ to $\overline{AB}$ is: \[ \frac{2}{6} \cdot 9 = 3 \] The area of triangle $ABE$ is: \[ \frac{1}{2} \cdot 2 \cdot 3 = 3 \] Conclusion: \[ \boxed{3} \] 12. Finding the height of the rectangular prism Given $144$ unit cubes form a rectangular prism and the perimeter of the base is $54$ units. Let the dimensions of the base be $a$ and $b$, and the height be $h$. The perimeter of the base is: \[ 2a + 2b = 54 \implies a + b = 27 \] The volume of the prism is: \[ a \cdot b \cdot h = 144 \] Substituting $b = 27 - a$: \[ a(27 - a)h = 144 \] We need to find integer solutions. Trying $a = 24$ and $b = 3$: \[ 24 \cdot 3 \cdot h = 144 \implies h = 2 \] Conclusion: \[ \boxed{2} \]	2	Other	math-word-problem	Yes	Yes	aops_forum	false
[u]Round 1[/u] [b]p1.[/b] Every angle of a regular polygon has degree measure $179.99$ degrees. How many sides does it have? [b]p2.[/b] What is $\frac{1}{20} + \frac{1}{1}+ \frac{1}{5}$ ? [b]p3.[/b] If the area bounded by the lines $y = 0$, $x = 0$, and $x = 3$ and the curve $y = f(x)$ is $10$ units, what is the area bounded by $y = 0$, $x = 0$, $x = 6$, and $y = f(x/2)$? [u]Round 2[/u] [b]p4.[/b] How many ways can $42$ be expressed as the sum of $2$ or more consecutive positive integers? [b]p5.[/b] How many integers less than or equal to $2015$ can be expressed as the sum of $2$ (not necessarily distinct) powers of two? [b]p6.[/b] $p,q$, and $q^2 - p^2$ are all prime. What is $pq$? [u]Round 3[/u] [b]p7.[/b] Let $p(x) = x^2 + ax + a$ be a polynomial with integer roots, where $a$ is an integer. What are all the possible values of $a$? [b]p8.[/b] In a given right triangle, the perimeter is $30$ and the sum of the squares of the sides is $338$. Find the lengths of the three sides. [b]p9.[/b] Each of the $6$ main diagonals of a regular hexagon is drawn, resulting in $6$ triangles. Each of those triangles is then split into $4$ equilateral triangles by connecting the midpoints of the $3$ sides. How many triangles are in the resulting figure? [u]Round 4[/u] [b]p10.[/b] Let $f = 5x+3y$, where $x$ and $y$ are positive real numbers such that $xy$ is $100$. Find the minimum possible value of $f$. [b]p11.[/b] An integer is called "Awesome" if its base $8$ expression contains the digit string $17$ at any point (i.e. if it ever has a $1$ followed immediately by a $7$). How many integers from $1$ to $500$ (base $10$) inclusive are Awesome? [b]p12.[/b] A certain pool table is a rectangle measuring $15 \times 24$ feet, with $4$ holes, one at each vertex. When playing pool, Joe decides that a ball has to hit at least $2$ sides before getting into a hole or else the shot does not count. What is the minimum distance a ball can travel after being hit on this table if it was hit at a vertex (assume it only stops after going into a hole) such that the shot counts? PS. You should use hide for answers. Rounds 5-8 have been posted [url=https://artofproblemsolving.com/community/c3h3157013p28696685]here [/url] and 9-12 [url=https://artofproblemsolving.com/community/c3h3158564p28715928]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. We are given that $ p, q $, and $ q^2 - p^2 $ are all prime numbers. We need to find the value of $ pq $. 2. First, consider the parity of $ p $ and $ q $. If both $ p $ and $ q $ are odd, then $ q^2 - p^2 $ will be even. The only even prime number is 2. Therefore, we would have: \[ q^2 - p^2 = 2 \] However, there are no two odd primes $ p $ and $ q $ such that $ q^2 - p^2 = 2 $. This leads to a contradiction. 3. If both $ p $ and $ q $ are even, then they must both be equal to 2 (since 2 is the only even prime). But then: \[ q^2 - p^2 = 2^2 - 2^2 = 4 - 4 = 0 \] which is not a prime number. This also leads to a contradiction. 4. Therefore, one of $ p $ or $ q $ must be 2 (the only even prime), and the other must be an odd prime. Without loss of generality, let $ p = 2 $. Then: \[ q^2 - p^2 = q^2 - 2^2 = q^2 - 4 \] Since $ q^2 - 4 $ is prime, we can write: \[ q^2 - 4 = (q - 2)(q + 2) \] For this product to be prime, one of the factors must be 1 (since the only way to get a prime number from a product of two integers is if one of them is 1). Therefore, we have: \[ q - 2 = 1 \quad \text{or} \quad q + 2 = 1 \] Solving these, we get: \[ q - 2 = 1 \implies q = 3 \] \[ q + 2 = 1 \implies q = -1 \quad (\text{not a prime}) \] Thus, the only solution is $ q = 3 $. 5. Finally, we calculate $ pq $: \[ pq = 2 \cdot 3 = 6 \] The final answer is $ \boxed{6} $.	6	Other	math-word-problem	Yes	Yes	aops_forum	false
[u]Round 5[/u] [b]p13.[/b] Sally is at the special glasses shop, where there are many different optical lenses that distort what she sees and cause her to see things strangely. Whenever she looks at a shape through lens $A$, she sees a shape with $2$ more sides than the original (so a square would look like a hexagon). When she looks through lens $B$, she sees the shape with $3$ fewer sides (so a hexagon would look like a triangle). How many sides are in the shape that has $200$ more diagonals when looked at from lense $A$ than from lense $B$? [b]p14.[/b] How many ways can you choose $2$ cells of a $5$ by $5$ grid such that they aren't in the same row or column? [b]p15.[/b] If $a + \frac{1}{b} = (2015)^{-1}$ and $b + \frac{1}{a} = (2016)^2$ then what are all the possible values of $b$? [u]Round 6[/u] [b]p16.[/b] In Canadian football, linebackers must wear jersey numbers from $30 -35$ while defensive linemen must wear numbers from $33 -38$ (both intervals are inclusive). If a team has $5$ linebackers and $4$ defensive linemen, how many ways can it assign jersey numbers to the $9$ players such that no two people have the same jersey number? [b]p17.[/b] What is the maximum possible area of a right triangle with hypotenuse $8$? [b]p18.[/b] $9$ people are to play touch football. One will be designated the quarterback, while the other eight will be divided into two (indistinct) teams of $4$. How many ways are there for this to be done? [u]Round 7[/u] [b]p19.[/b] Express the decimal $0.3$ in base $7$. [b]p20.[/b] $2015$ people throw their hats in a pile. One at a time, they each take one hat out of the pile so that each has a random hat. What is the expected number of people who get their own hat? [b]p21.[/b] What is the area of the largest possible trapezoid that can be inscribed in a semicircle of radius $4$? [u]Round 8[/u] [b]p22.[/b] What is the base $7$ expression of $1211_3 \cdot 1110_2 \cdot 292_{11} \cdot 20_3$ ? [b]p23.[/b] Let $f(x)$ equal the ratio of the surface area of a sphere of radius $x$ to the volume of that same sphere. Let $g(x)$ be a quadratic polynomial in the form $x^2 + bx + c$ with $g(6) = 0$ and the minimum value of $g(x)$ equal to $c$. Express $g(x)$ as a function of $f(x)$ (e.g. in terms of $f(x)$). [b]p24.[/b] In the country of Tahksess, the income tax code is very complicated. Citizens are taxed $40\%$ on their first $\$20, 000$ and $45\%$ on their next $\$40, 000$ and $50\%$ on their next $\$60, 000$ and so on, with each $5\%$ increase in tax rate aecting $\$20, 000$ more than the previous tax rate. The maximum tax rate, however, is $90\%$. What is the overall tax rate (percentage of money owed) on $1$ million dollars in income? PS. You should use hide for answers. Rounds 1-4 have been posted [url=https://artofproblemsolving.com/community/c3h3157009p28696627]here [/url] and 9-12 [url=https://artofproblemsolving.com/community/c3h3158564p28715928]here[/url]. .Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. Let the people be $ P_1, P_2, P_3, \ldots, P_{2015} $, who retrieve their hats in that order. 2. Define $ x_1, x_2, x_3, \ldots, x_{2015} $ such that, for all $ i \in \{1,2,3,\ldots,2015\} $, $ x_i=0 $ if $ P_i $ does not get their hat and $ x_i=1 $ if $ P_i $ does get their hat. 3. We wish to find $ \mathbb{E}\left[\sum_{i=1}^{2015} x_i \right] $. By the Linearity of Expectation, this is $ \sum_{i=1}^{2015} \mathbb{E}[x_i] $. 4. Note that there is a $ \frac{i-1}{2015} $ probability that $ P_i $'s hat is chosen by one of the first $ i-1 $ people and a $ \frac{2015-(i-1)}{2015} $ probability that $ P_i $'s hat is not chosen by any of the first $ i-1 $ people. 5. If $ P_i $'s hat is chosen by one of the first $ i-1 $ people, then there is a probability of $ 0 $ that $ P_i $ will get his own hat. If $ P_i $'s hat is not chosen by any of the first $ i-1 $ people, then one of the remaining $ 2015-(i-1) $ hats is his, so there is a $ \frac{1}{2015-(i-1)} $ probability that he gets it. 6. Hence, the probability that $ P_i $ gets his own hat is \[ \frac{i-1}{2015} \cdot 0 + \frac{2015-(i-1)}{2015} \cdot \frac{1}{2015-(i-1)} = 0 + \frac{1}{2015} = \frac{1}{2015}, \] so \[ \mathbb{E}[x_i] = \frac{1}{2015} \quad \forall \quad i \in \{1,2,3,\ldots,2015\}. \] 7. Therefore, the expected number of people who get their own hat is \[ \sum_{i=1}^{2015} \mathbb{E}[x_i] = \sum_{i=1}^{2015} \frac{1}{2015} = 2015 \cdot \frac{1}{2015} = 1. \] The final answer is $\boxed{1}$	1	Logic and Puzzles	math-word-problem	Yes	Yes	aops_forum	false
Albert rolls a fair six-sided die thirteen times. For each time he rolls a number that is strictly greater than the previous number he rolled, he gains a point, where his first roll does not gain him a point. Find the expected number of points that Albert receives. [i]Proposed by Nathan Ramesh	1. Define the problem: Albert rolls a fair six-sided die thirteen times. For each roll after the first, he gains a point if the number rolled is strictly greater than the previous number. We need to find the expected number of points Albert receives. 2. Calculate the probability of gaining a point on a single roll: - Let $X_i$ be the outcome of the $i$-th roll. - For $i \geq 2$, Albert gains a point if $X_i > X_{i-1}$. - The probability that $X_i > X_{i-1}$ can be calculated by considering all possible outcomes of $X_{i-1}$ and $X_i$: \[ P(X_i > X_{i-1}) = \frac{1}{6} \cdot \frac{5}{6} + \frac{1}{6} \cdot \frac{4}{6} + \frac{1}{6} \cdot \frac{3}{6} + \frac{1}{6} \cdot \frac{2}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot 0 \] \[ = \frac{1}{6} \left( \frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6} + 0 \right) \] \[ = \frac{1}{6} \cdot \frac{15}{6} = \frac{15}{36} = \frac{5}{12} \] 3. Calculate the expected number of points: - Since there are 12 comparisons (from the 2nd roll to the 13th roll), and each comparison has an expected value of $\frac{5}{12}$ points: \[ E(\text{total points}) = 12 \cdot \frac{5}{12} = 5 \] The final answer is $\boxed{5}$.	5	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
A single elimination tournament is held with $2016$ participants. In each round, players pair up to play games with each other. There are no ties, and if there are an odd number of players remaining before a round then one person will get a bye for the round. Find the minimum number of rounds needed to determine a winner. [i]Proposed by Nathan Ramesh	1. In a single elimination tournament, each round halves the number of participants (or nearly halves if there is an odd number of participants, in which case one participant gets a bye). 2. We start with 2016 participants. After each round, the number of participants is approximately halved. Let's calculate the number of participants after each round: - After 1st round: $\left\lceil \frac{2016}{2} \right\rceil = 1008$ - After 2nd round: $\left\lceil \frac{1008}{2} \right\rceil = 504$ - After 3rd round: $\left\lceil \frac{504}{2} \right\rceil = 252$ - After 4th round: $\left\lceil \frac{252}{2} \right\rceil = 126$ - After 5th round: $\left\lceil \frac{126}{2} \right\rceil = 63$ At this point, we have 63 participants, which is an odd number. One participant will get a bye in the next round: - After 6th round: $\left\lceil \frac{63 + 1}{2} \right\rceil = 32$ Continuing with the halving process: - After 7th round: $\left\lceil \frac{32}{2} \right\rceil = 16$ - After 8th round: $\left\lceil \frac{16}{2} \right\rceil = 8$ - After 9th round: $\left\lceil \frac{8}{2} \right\rceil = 4$ - After 10th round: $\left\lceil \frac{4}{2} \right\rceil = 2$ - After 11th round: $\left\lceil \frac{2}{2} \right\rceil = 1$ Thus, it takes 11 rounds to determine a winner. The final answer is $\boxed{11}$	11	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
[b]p1.[/b] Compute $2020 \cdot \left( 2^{(0\cdot1)} + 9 - \frac{(20^1)}{8}\right)$. [b]p2.[/b] Nathan has five distinct shirts, three distinct pairs of pants, and four distinct pairs of shoes. If an “outfit” has a shirt, pair of pants, and a pair of shoes, how many distinct outfits can Nathan make? [b]p3.[/b] Let $ABCD$ be a rhombus such that $\vartriangle ABD$ and $\vartriangle BCD$ are equilateral triangles. Find the angle measure of $\angle ACD$ in degrees. [b]p4.[/b] Find the units digit of $2019^{2019}$. [b]p5.[/b] Determine the number of ways to color the four vertices of a square red, white, or blue if two colorings that can be turned into each other by rotations and reflections are considered the same. [b]p6.[/b] Kathy rolls two fair dice numbered from $1$ to $6$. At least one of them comes up as a $4$ or $5$. Compute the probability that the sumof the numbers of the two dice is at least $10$. [b]p7.[/b] Find the number of ordered pairs of positive integers $(x, y)$ such that $20x +19y = 2019$. [b]p8.[/b] Let $p$ be a prime number such that both $2p -1$ and $10p -1$ are prime numbers. Find the sum of all possible values of $p$. [b]p9.[/b] In a square $ABCD$ with side length $10$, let $E$ be the intersection of $AC$ and $BD$. There is a circle inscribed in triangle $ABE$ with radius $r$ and a circle circumscribed around triangle $ABE$ with radius $R$. Compute $R -r$ . [b]p10.[/b] The fraction $\frac{13}{37 \cdot 77}$ can be written as a repeating decimal $0.a_1a_2...a_{n-1}a_n$ with $n$ digits in its shortest repeating decimal representation. Find $a_1 +a_2 +...+a_{n-1}+a_n$. [b]p11.[/b] Let point $E$ be the midpoint of segment $AB$ of length $12$. Linda the ant is sitting at $A$. If there is a circle $O$ of radius $3$ centered at $E$, compute the length of the shortest path Linda can take from $A$ to $B$ if she can’t cross the circumference of $O$. [b]p12.[/b] Euhan and Minjune are playing tennis. The first one to reach $25$ points wins. Every point ends with Euhan calling the ball in or out. If the ball is called in, Minjune receives a point. If the ball is called out, Euhan receives a point. Euhan always makes the right call when the ball is out. However, he has a $\frac34$ chance of making the right call when the ball is in, meaning that he has a $\frac14$ chance of calling a ball out when it is in. The probability that the ball is in is equal to the probability that the ball is out. If Euhan won, determine the expected number of wrong callsmade by Euhan. [b]p13.[/b] Find the number of subsets of $\{1, 2, 3, 4, 5, 6,7\}$ which contain four consecutive numbers. [b]p14.[/b] Ezra and Richard are playing a game which consists of a series of rounds. In each round, one of either Ezra or Richard receives a point. When one of either Ezra or Richard has three more points than the other, he is declared the winner. Find the number of games which last eleven rounds. Two games are considered distinct if there exists a round in which the two games had different outcomes. [b]p15.[/b] There are $10$ distinct subway lines in Boston, each of which consists of a path of stations. Using any $9$ lines, any pair of stations are connected. However, among any $8$ lines there exists a pair of stations that cannot be reached from one another. It happens that the number of stations is minimized so this property is satisfied. What is the average number of stations that each line passes through? [b]p16.[/b] There exist positive integers $k$ and $3\nmid m$ for which $$1 -\frac12 + \frac13 - \frac14 +...+ \frac{1}{53}-\frac{1}{54}+\frac{1}{55}=\frac{3^k \times m}{28\times 29\times ... \times 54\times 55}.$$ Find the value $k$. [b]p17.[/b] Geronimo the giraffe is removing pellets from a box without replacement. There are $5$ red pellets, $10$ blue pellets, and $15$ white pellets. Determine the probability that all of the red pellets are removed before all the blue pellets and before all of the white pellets are removed. [b]p18.[/b] Find the remainder when $$70! \left( \frac{1}{4 \times 67}+ \frac{1}{5 \times 66}+...+ \frac{1}{66\times 5}+ \frac{1}{67\times 4} \right)$$ is divided by $71$. [b]p19.[/b] Let $A_1A_2...A_{12}$ be the regular dodecagon. Let $X$ be the intersection of $A_1A_2$ and $A_5A_{11}$. Given that $X A_2 \cdot A_1A_2 = 10$, find the area of dodecagon. [b]p20.[/b] Evaluate the following infinite series: $$\sum^{\infty}_{n=1}\sum^{\infty}_{m=1} \frac{n \sec^2m -m \tan^2 n}{3^{m+n}(m+n)}$$. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	4. Find the units digit of $2019^{2019}$. Step-by-step solution: 1. Determine the units digit of $2019$, which is $9$. 2. Observe the pattern in the units digits of powers of $9$: \[ 9^1 = 9, \quad 9^2 = 81, \quad 9^3 = 729, \quad 9^4 = 6561, \quad \ldots \] The units digits repeat every 2 powers: $9, 1, 9, 1, \ldots$. 3. Since $2019$ is odd, the units digit of $2019^{2019}$ is the same as the units digit of $9^1$, which is $9$. The final answer is $\boxed{9}$.	9	Other	math-word-problem	Yes	Yes	aops_forum	false
Compute the maximum integer value of $k$ such that $2^k$ divides $3^{2n+3}+40n-27$ for any positive integer $n$.	To find the maximum integer value of $ k $ such that $ 2^k $ divides $ 3^{2n+3} + 40n - 27 $ for any positive integer $ n $, we will use modular arithmetic and analyze the expression modulo powers of 2. 1. Rewrite the expression: \[ 3^{2n+3} + 40n - 27 \] We can factor out common terms: \[ 3^{2n+3} = 27 \cdot 9^n \] So the expression becomes: \[ 27 \cdot 9^n + 40n - 27 \] Simplify: \[ 27(9^n - 1) + 40n \] 2. Analyze modulo $ 2^k $: We need to find the highest $ k $ such that: \[ 27(9^n - 1) + 40n \equiv 0 \pmod{2^k} \] 3. Check modulo $ 2 $: \[ 27 \equiv 1 \pmod{2} \] \[ 9 \equiv 1 \pmod{2} \] \[ 27(9^n - 1) + 40n \equiv 1(1^n - 1) + 0 \equiv 0 \pmod{2} \] This is true for all $ n $. 4. Check modulo $ 4 $: \[ 27 \equiv 3 \pmod{4} \] \[ 9 \equiv 1 \pmod{4} \] \[ 27(9^n - 1) + 40n \equiv 3(1^n - 1) + 0 \equiv 0 \pmod{4} \] This is true for all $ n $. 5. Check modulo $ 8 $: \[ 27 \equiv 3 \pmod{8} \] \[ 9 \equiv 1 \pmod{8} \] \[ 27(9^n - 1) + 40n \equiv 3(1^n - 1) + 0 \equiv 0 \pmod{8} \] This is true for all $ n $. 6. Check modulo $ 16 $: \[ 27 \equiv 11 \pmod{16} \] \[ 9 \equiv 9 \pmod{16} \] \[ 27(9^n - 1) + 40n \equiv 11(9^n - 1) + 8n \pmod{16} \] For $ n = 1 $: \[ 11(9 - 1) + 8 \equiv 11 \cdot 8 + 8 \equiv 88 + 8 \equiv 96 \equiv 0 \pmod{16} \] For $ n = 2 $: \[ 11(81 - 1) + 16 \equiv 11 \cdot 80 + 16 \equiv 880 + 16 \equiv 896 \equiv 0 \pmod{16} \] This is true for all $ n $. 7. Check modulo $ 32 $: \[ 27 \equiv 27 \pmod{32} \] \[ 9 \equiv 9 \pmod{32} \] \[ 27(9^n - 1) + 40n \equiv 27(9^n - 1) + 8n \pmod{32} \] For $ n = 1 $: \[ 27(9 - 1) + 40 \equiv 27 \cdot 8 + 40 \equiv 216 + 40 \equiv 256 \equiv 0 \pmod{32} \] For $ n = 2 $: \[ 27(81 - 1) + 80 \equiv 27 \cdot 80 + 80 \equiv 2160 + 80 \equiv 2240 \equiv 0 \pmod{32} \] This is true for all $ n $. 8. Check modulo $ 64 $: \[ 27 \equiv 27 \pmod{64} \] \[ 9 \equiv 9 \pmod{64} \] \[ 27(9^n - 1) + 40n \equiv 27(9^n - 1) + 40n \pmod{64} \] For $ n = 1 $: \[ 27(9 - 1) + 40 \equiv 27 \cdot 8 + 40 \equiv 216 + 40 \equiv 256 \equiv 0 \pmod{64} \] For $ n = 2 $: \[ 27(81 - 1) + 80 \equiv 27 \cdot 80 + 80 \equiv 2160 + 80 \equiv 2240 \equiv 0 \pmod{64} \] This is true for all $ n $. Thus, the maximum $ k $ such that $ 2^k $ divides $ 3^{2n+3} + 40n - 27 $ for any positive integer $ n $ is $ k = 6 $. The final answer is $ \boxed{6} $.	6	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Farmer Boso has a busy farm with lots of animals. He tends to $5b$ cows, $5a +7$ chickens, and $b^{a-5}$ insects. Note that each insect has $6$ legs. The number of cows is equal to the number of insects. The total number of legs present amongst his animals can be expressed as $\overline{LLL }+1$, where $L$ stands for a digit. Find $L$.	1. We are given that the number of cows is equal to the number of insects. This can be expressed as: \[ 5b = b^{a-5} \] Dividing both sides by $ b $ (assuming $ b \neq 0 $): \[ 5 = b^{a-6} \] Since $ a $ and $ b $ are integers, we need to find integer values of $ a $ and $ b $ that satisfy this equation. 2. We test possible values of $ b $: - If $ b = 5 $: \[ 5 = 5^{a-6} \] This implies: \[ 5 = 5^{a-6} \implies a-6 = 1 \implies a = 7 \] Therefore, $ b = 5 $ and $ a = 7 $ are solutions. 3. Now, we calculate the number of each type of animal: - Number of cows: \[ 5b = 5 \cdot 5 = 25 \] - Number of chickens: \[ 5a + 7 = 5 \cdot 7 + 7 = 35 + 7 = 42 \] - Number of insects: \[ b^{a-5} = 5^{7-5} = 5^2 = 25 \] 4. Next, we calculate the total number of legs: - Cows have 4 legs each: \[ 25 \text{ cows} \times 4 \text{ legs/cow} = 100 \text{ legs} \] - Chickens have 2 legs each: \[ 42 \text{ chickens} \times 2 \text{ legs/chicken} = 84 \text{ legs} \] - Insects have 6 legs each: \[ 25 \text{ insects} \times 6 \text{ legs/insect} = 150 \text{ legs} \] 5. Summing up all the legs: \[ 100 + 84 + 150 = 334 \] We are given that the total number of legs can be expressed as $ \overline{LLL} + 1 $. Therefore: \[ \overline{LLL} + 1 = 334 \implies \overline{LLL} = 333 \] 6. From this, we can deduce that $ L = 3 $. The final answer is $ \boxed{3} $	3	Algebra	math-word-problem	Yes	Yes	aops_forum	false
Find the greatest possible distance between any two points inside or along the perimeter of an equilateral triangle with side length $2$. [i]Proposed by Alex Li[/i]	1. Consider an equilateral triangle with side length $2$. Let's denote the vertices of the triangle as $A$, $B$, and $C$. 2. The distance between any two vertices of an equilateral triangle is equal to the side length of the triangle. Therefore, the distance between any two vertices $A$ and $B$, $B$ and $C$, or $C$ and $A$ is $2$. 3. To confirm that this is the greatest possible distance, we need to consider any two points inside or along the perimeter of the triangle. The maximum distance between any two points inside or along the perimeter of the triangle cannot exceed the distance between the vertices, as the vertices are the farthest apart. 4. Therefore, the greatest possible distance between any two points inside or along the perimeter of an equilateral triangle with side length $2$ is the distance between two vertices, which is $2$. \[ \boxed{2} \]	2	Geometry	math-word-problem	Yes	Yes	aops_forum	false
Aidan rolls a pair of fair, six sided dice. Let$ n$ be the probability that the product of the two numbers at the top is prime. Given that $n$ can be written as $a/b$ , where $a$ and $b$ are relatively prime positive integers, find $a +b$. [i]Proposed by Aidan Duncan[/i]	1. To determine the probability that the product of the two numbers on the top of the dice is prime, we first need to understand the conditions under which the product of two numbers is prime. A product of two numbers is prime if and only if one of the numbers is 1 and the other number is a prime number. This is because a prime number has exactly two distinct positive divisors: 1 and itself. 2. The prime numbers less than or equal to 6 (the numbers on a six-sided die) are 2, 3, and 5. Therefore, the possible prime numbers that can appear on the dice are 2, 3, and 5. 3. For each prime number, there are two possible scenarios: the prime number can appear on the first die and 1 on the second die, or the prime number can appear on the second die and 1 on the first die. This gives us a total of 2 choices for each prime number. 4. Since there are 3 prime numbers (2, 3, and 5), and each prime number can appear in 2 different ways (as described in step 3), the total number of favorable outcomes is: \[ 2 \times 3 = 6 \] 5. The total number of possible outcomes when rolling two six-sided dice is: \[ 6 \times 6 = 36 \] 6. Therefore, the probability $ n $ that the product of the two numbers is prime is: \[ n = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{6}{36} = \frac{1}{6} \] 7. Given that $ n $ can be written as $ \frac{a}{b} $, where $ a $ and $ b $ are relatively prime positive integers, we have $ a = 1 $ and $ b = 6 $. 8. The problem asks for $ a + b $: \[ a + b = 1 + 6 = 7 \] The final answer is $ \boxed{7} $	7	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Find the number of arithmetic sequences $a_1,a_2,a_3$ of three nonzero integers such that the sum of the terms in the sequence is equal to the product of the terms in the sequence. [i]Proposed by Sammy Charney[/i]	1. Let $ a_1 = x $, $ a_2 = x + d $, and $ a_3 = x + 2d $, where $ x, d \in \mathbb{Z} $. Given that the sum of the terms in the sequence is equal to the product of the terms in the sequence, we have: \[ a_1 + a_2 + a_3 = a_1 a_2 a_3 \] Substituting the values of $ a_1 $, $ a_2 $, and $ a_3 $, we get: \[ x + (x + d) + (x + 2d) = x (x + d) (x + 2d) \] Simplifying the left-hand side: \[ 3x + 3d = x (x + d) (x + 2d) \] Since $ x + d \neq 0 $, we can divide both sides by $ x + d $: \[ 3 = x (x + 2d) \] Expanding and rearranging the equation: \[ x^2 + 2dx - 3 = 0 \] 2. Solving the quadratic equation $ x^2 + 2dx - 3 = 0 $ using the quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = 2d $, and $ c = -3 $: \[ x = \frac{-2d \pm \sqrt{(2d)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \] Simplifying under the square root: \[ x = \frac{-2d \pm \sqrt{4d^2 + 12}}{2} \] \[ x = \frac{-2d \pm 2\sqrt{d^2 + 3}}{2} \] \[ x = -d \pm \sqrt{d^2 + 3} \] 3. For $ x $ to be an integer, $ \sqrt{d^2 + 3} $ must be an integer. Let $ k = \sqrt{d^2 + 3} $, then: \[ k^2 = d^2 + 3 \] \[ k^2 - d^2 = 3 \] \[ (k - d)(k + d) = 3 \] The pairs $(k - d, k + d)$ that satisfy this equation are $(1, 3)$ and $(-1, -3)$. Solving these pairs: - For $ k - d = 1 $ and $ k + d = 3 $: \[ k = 2, \quad d = 1 \] - For $ k - d = -1 $ and $ k + d = -3 $: \[ k = -2, \quad d = -1 \] 4. Substituting $ d = 1 $ and $ d = -1 $ back into the equation for $ x $: - For $ d = 1 $: \[ x = -1 \pm 2 \] \[ x = 1 \quad \text{or} \quad x = -3 \] - For $ d = -1 $: \[ x = 1 \pm 2 \] \[ x = 3 \quad \text{or} \quad x = -1 \] 5. The pairs $(x, d)$ are $(1, 1)$, $(-3, 1)$, $(3, -1)$, and $(-1, -1)$. These yield the sequences: - For $(1, 1)$: $ 1, 2, 3 $ - For $(-3, 1)$: $ -3, -2, -1 $ - For $(3, -1)$: $ 3, 2, 1 $ - For $(-1, -1)$: $ -1, -2, -3 $ Hence, there are a total of $ \boxed{4} $ arithmetic sequences of three nonzero integers such that the sum of the terms in the sequence is equal to the product of the terms in the sequence.	4	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Let $f (x)$ be a function mapping real numbers to real numbers. Given that $f (f (x)) =\frac{1}{3x}$, and $f (2) =\frac19$, find $ f\left(\frac{1}{6}\right)$. [i]Proposed by Zachary Perry[/i]	1. We start with the given functional equation: \[ f(f(x)) = \frac{1}{3x} \] and the specific value: \[ f(2) = \frac{1}{9} \] 2. Substitute $ x = 2 $ into the functional equation: \[ f(f(2)) = \frac{1}{3 \cdot 2} = \frac{1}{6} \] Since $ f(2) = \frac{1}{9} $, we have: \[ f\left(\frac{1}{9}\right) = \frac{1}{6} \] 3. Next, we need to find $ f\left(\frac{1}{6}\right) $. Using the functional equation again, we substitute $ x = \frac{1}{9} $: \[ f\left(f\left(\frac{1}{9}\right)\right) = \frac{1}{3 \cdot \frac{1}{9}} = 3 \] We already know that $ f\left(\frac{1}{9}\right) = \frac{1}{6} $, so: \[ f\left(\frac{1}{6}\right) = 3 \] Thus, we have determined that: \[ f\left(\frac{1}{6}\right) = 3 \] The final answer is $\boxed{3}$	3	Other	math-word-problem	Yes	Yes	aops_forum	false
Kevin colors a ninja star on a piece of graph paper where each small square has area $1$ square inch. Find the area of the region colored, in square inches. [img]https://cdn.artofproblemsolving.com/attachments/3/3/86f0ae7465e99d3e4bd3a816201383b98dc429.png[/img]	1. Identify the outer triangles: The ninja star can be divided into 4 outer triangles. Each of these triangles has a base of 2 inches and a height of 2 inches. 2. Calculate the area of one outer triangle: The area $ A $ of a triangle is given by: \[ A = \frac{1}{2} \times \text{base} \times \text{height} \] For one outer triangle: \[ A = \frac{1}{2} \times 2 \times 2 = \frac{4}{2} = 2 \text{ square inches} \] 3. Calculate the total area of the outer triangles: Since there are 4 outer triangles: \[ \text{Total area of outer triangles} = 4 \times 2 = 8 \text{ square inches} \] 4. Identify the inner triangles: The ninja star also contains 4 inner triangles. Each of these triangles has a base of 2 inches and a height of 1 inch. 5. Calculate the area of one inner triangle: For one inner triangle: \[ A = \frac{1}{2} \times 2 \times 1 = \frac{2}{2} = 1 \text{ square inch} \] 6. Calculate the total area of the inner triangles: Since there are 4 inner triangles: \[ \text{Total area of inner triangles} = 4 \times 1 = 4 \text{ square inches} \] 7. Calculate the total area of the colored region: Adding the areas of the outer and inner triangles: \[ \text{Total area} = 8 + 4 = 12 \text{ square inches} \] The final answer is $\boxed{12}$ square inches.	12	Geometry	math-word-problem	Yes	Yes	aops_forum	false
The equation of line $\ell_1$ is $24x-7y = 319$ and the equation of line $\ell_2$ is $12x-5y = 125$. Let $a$ be the number of positive integer values $n$ less than $2023$ such that for both $\ell_1$ and $\ell_2$ there exists a lattice point on that line that is a distance of $n$ from the point $(20,23)$. Determine $a$. [i]Proposed by Christopher Cheng[/i] [hide=Solution][i]Solution. [/i] $\boxed{6}$ Note that $(20,23)$ is the intersection of the lines $\ell_1$ and $\ell_2$. Thus, we only care about lattice points on the the two lines that are an integer distance away from $(20,23)$. Notice that $7$ and $24$ are part of the Pythagorean triple $(7,24,25)$ and $5$ and $12$ are part of the Pythagorean triple $(5,12,13)$. Thus, points on $\ell_1$ only satisfy the conditions when $n$ is divisible by $25$ and points on $\ell_2$ only satisfy the conditions when $n$ is divisible by $13$. Therefore, $a$ is just the number of positive integers less than $2023$ that are divisible by both $25$ and $13$. The LCM of $25$ and $13$ is $325$, so the answer is $\boxed{6}$.[/hide]	1. Intersection of Lines: We start by finding the intersection of the lines $\ell_1$ and $\ell_2$. The equations are: \[ 24x - 7y = 319 \] \[ 12x - 5y = 125 \] To find the intersection, we solve this system of linear equations. We can use the method of elimination or substitution. Here, we will use elimination. 2. Elimination Method: Multiply the second equation by 2 to align the coefficients of $x$: \[ 24x - 10y = 250 \] Now subtract the second equation from the first: \[ (24x - 7y) - (24x - 10y) = 319 - 250 \] Simplifying, we get: \[ 3y = 69 \implies y = 23 \] Substitute $y = 23$ back into the first equation: \[ 24x - 7(23) = 319 \implies 24x - 161 = 319 \implies 24x = 480 \implies x = 20 \] Therefore, the intersection point is $(20, 23)$. 3. Distance from $(20, 23)$: We need to find lattice points on $\ell_1$ and $\ell_2$ that are an integer distance $n$ from $(20, 23)$. The distance formula is: \[ d = \sqrt{(x - 20)^2 + (y - 23)^2} \] For $d$ to be an integer, $(x - 20)^2 + (y - 23)^2$ must be a perfect square, say $k^2$. Thus: \[ (x - 20)^2 + (y - 23)^2 = n^2 \] 4. Lattice Points on $\ell_1$: The line $\ell_1$ can be rewritten in slope-intercept form: \[ y = \frac{24}{7}x - \frac{319}{7} \] Notice that $24$ and $7$ are part of the Pythagorean triple $(7, 24, 25)$. Therefore, the distance $n$ must be a multiple of $25$ for there to be lattice points on $\ell_1$. 5. Lattice Points on $\ell_2$: The line $\ell_2$ can be rewritten in slope-intercept form: \[ y = \frac{12}{5}x - \frac{125}{5} \] Notice that $12$ and $5$ are part of the Pythagorean triple $(5, 12, 13)$. Therefore, the distance $n$ must be a multiple of $13$ for there to be lattice points on $\ell_2$. 6. Common Multiples: We need $n$ to be a multiple of both $25$ and $13$. The least common multiple (LCM) of $25$ and $13$ is: \[ \text{LCM}(25, 13) = 325 \] We need to count the multiples of $325$ that are less than $2023$: \[ \left\lfloor \frac{2023}{325} \right\rfloor = 6 \] Conclusion: The number of positive integer values $n$ less than $2023$ such that there exists a lattice point on both $\ell_1$ and $\ell_2$ at a distance $n$ from $(20, 23)$ is $\boxed{6}$.	6	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Eddie has a study block that lasts $1$ hour. It takes Eddie $25$ minutes to do his homework and $5$ minutes to play a game of Clash Royale. He can’t do both at the same time. How many games can he play in this study block while still completing his homework? [i]Proposed by Edwin Zhao[/i] [hide=Solution] [i]Solution.[/i] $\boxed{7}$ Study block lasts 60 minutes, thus he has 35 minutes to play Clash Royale, during which he can play $\frac{35}{5}=\boxed{7}$ games. [/hide]	1. Convert the study block duration from hours to minutes: \[ 1 \text{ hour} = 60 \text{ minutes} \] 2. Subtract the time taken to do homework from the total study block time: \[ 60 \text{ minutes} - 25 \text{ minutes} = 35 \text{ minutes} \] 3. Determine the number of games Eddie can play in the remaining time. Each game takes 5 minutes: \[ \frac{35 \text{ minutes}}{5 \text{ minutes/game}} = 7 \text{ games} \] Conclusion: Eddie can play 7 games of Clash Royale while still completing his homework within the 1-hour study block. The final answer is $\boxed{7}$.	7	Logic and Puzzles	math-word-problem	Yes	Yes	aops_forum	false
Sam Wang decides to evaluate an expression of the form $x +2 \cdot 2+ y$. However, he unfortunately reads each ’plus’ as a ’times’ and reads each ’times’ as a ’plus’. Surprisingly, he still gets the problem correct. Find $x + y$. [i]Proposed by Edwin Zhao[/i] [hide=Solution] [i]Solution.[/i] $\boxed{4}$ We have $x+2*2+y=x \cdot 2+2 \cdot y$. When simplifying, we have $x+y+4=2x+2y$, and $x+y=4$. [/hide]	1. The original expression given is $ x + 2 \cdot 2 + y $. 2. Sam reads each 'plus' as 'times' and each 'times' as 'plus'. Therefore, the expression Sam evaluates is $ x \cdot 2 + 2 + y $. 3. We need to simplify both expressions and set them equal to each other since Sam gets the problem correct. 4. Simplify the original expression: \[ x + 2 \cdot 2 + y = x + 4 + y \] 5. Simplify the expression Sam evaluates: \[ x \cdot 2 + 2 + y = 2x + 2 + y \] 6. Set the simplified expressions equal to each other: \[ x + 4 + y = 2x + 2 + y \] 7. Subtract $ y $ from both sides: \[ x + 4 = 2x + 2 \] 8. Subtract $ x $ from both sides: \[ 4 = x + 2 \] 9. Subtract 2 from both sides: \[ 2 = x \] 10. Substitute $ x = 2 $ back into the equation $ x + y = 4 $: \[ 2 + y = 4 \] 11. Solve for $ y $: \[ y = 2 \] 12. Therefore, $ x + y = 2 + 2 = 4 $. The final answer is $ \boxed{4} $	4	Algebra	math-word-problem	Yes	Yes	aops_forum	false
Blue rolls a fair $n$-sided die that has sides its numbered with the integers from $1$ to $n$, and then he flips a coin. Blue knows that the coin is weighted to land heads either $\dfrac{1}{3}$ or $\dfrac{2}{3}$ of the time. Given that the probability of both rolling a $7$ and flipping heads is $\dfrac{1}{15}$, find $n$. [i]Proposed by Jacob Xu[/i] [hide=Solution][i]Solution[/i]. $\boxed{10}$ The chance of getting any given number is $\dfrac{1}{n}$ , so the probability of getting $7$ and heads is either $\dfrac{1}{n} \cdot \dfrac{1}{3}=\dfrac{1}{3n}$ or $\dfrac{1}{n} \cdot \dfrac{2}{3}=\dfrac{2}{3n}$. We get that either $n = 5$ or $n = 10$, but since rolling a $7$ is possible, only $n = \boxed{10}$ is a solution.[/hide]	1. Let's denote the probability of rolling any specific number on an $n$-sided die as $\frac{1}{n}$, since the die is fair. 2. The probability of flipping heads with the weighted coin is either $\frac{1}{3}$ or $\frac{2}{3}$. 3. We are given that the probability of both rolling a $7$ and flipping heads is $\frac{1}{15}$. We can set up the following equations based on the given probabilities: \[ \frac{1}{n} \cdot \frac{1}{3} = \frac{1}{15} \quad \text{or} \quad \frac{1}{n} \cdot \frac{2}{3} = \frac{1}{15} \] 4. Solving the first equation: \[ \frac{1}{n} \cdot \frac{1}{3} = \frac{1}{15} \] \[ \frac{1}{3n} = \frac{1}{15} \] \[ 3n = 15 \] \[ n = 5 \] 5. Solving the second equation: \[ \frac{1}{n} \cdot \frac{2}{3} = \frac{1}{15} \] \[ \frac{2}{3n} = \frac{1}{15} \] \[ 2 \cdot 15 = 3n \] \[ 30 = 3n \] \[ n = 10 \] 6. Since rolling a $7$ is only possible if $n \geq 7$, we discard $n = 5$ as a solution. Therefore, the only valid solution is $n = 10$. The final answer is $\boxed{10}$	10	Logic and Puzzles	math-word-problem	Yes	Yes	aops_forum	false
Find the least positive integer $k$ such that when $\frac{k}{2023}$ is written in simplest form, the sum of the numerator and denominator is divisible by $7$. [i]Proposed byMuztaba Syed[/i]	1. We need to find the least positive integer $ k $ such that when $\frac{k}{2023}$ is written in simplest form, the sum of the numerator and denominator is divisible by $ 7 $. 2. First, note that $ 2023 $ is not divisible by $ 7 $. We can verify this by checking the divisibility rule for $ 7 $: \[ 2023 \div 7 = 289 \quad \text{(not an integer)} \] Therefore, $ 2023 $ is not divisible by $ 7 $. 3. Let $ \frac{k}{2023} $ be written in simplest form as $ \frac{a}{b} $. Since $ 2023 $ is not divisible by $ 7 $, $ b $ must be $ 2023 $ and $ a $ must be $ k $ (since $ k $ and $ 2023 $ are coprime). 4. We need the sum of the numerator and denominator to be divisible by $ 7 $: \[ k + 2023 \equiv 0 \pmod{7} \] Simplifying, we find: \[ k \equiv -2023 \pmod{7} \] 5. Calculate $ 2023 \mod 7 $: \[ 2023 \div 7 = 289 \quad \text{(remainder 0)} \] Therefore: \[ 2023 \equiv 0 \pmod{7} \] So: \[ k \equiv 0 \pmod{7} \] 6. The smallest positive integer $ k $ that satisfies this congruence is $ k = 7 $. 7. Verify that $ \frac{7}{2023} $ is in simplest form: - The greatest common divisor (gcd) of $ 7 $ and $ 2023 $ is $ 1 $ (since $ 2023 $ is not divisible by $ 7 $). 8. Check the sum: \[ 7 + 2023 = 2030 \] Verify divisibility by $ 7 $: \[ 2030 \div 7 = 290 \quad \text{(an integer)} \] Thus, the least positive integer $ k $ such that the sum of the numerator and denominator of $\frac{k}{2023}$ in simplest form is divisible by $ 7 $ is $ k = 7 $. The final answer is $ \boxed{7} $.	7	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Given that the base-$17$ integer $\overline{8323a02421_{17}}$ (where a is a base-$17$ digit) is divisible by $\overline{16_{10}}$, find $a$. Express your answer in base $10$. [i]Proposed by Jonathan Liu[/i]	To determine the value of $ a $ such that the base-17 integer $\overline{8323a02421_{17}}$ is divisible by $16_{10}$, we can use the property that a number is divisible by 16 if the sum of its digits is divisible by 16. 1. Convert the base-17 number to a sum of its digits: \[ \overline{8323a02421_{17}} = 8_{17} + 3_{17} + 2_{17} + 3_{17} + a_{17} + 0_{17} + 2_{17} + 4_{17} + 2_{17} + 1_{17} \] 2. Sum the digits: \[ 8 + 3 + 2 + 3 + a + 0 + 2 + 4 + 2 + 1 = 25 + a \] 3. Find the modulo 16 of the sum: \[ 25 + a \equiv 9 + a \pmod{16} \] 4. Set up the congruence equation for divisibility by 16: \[ 9 + a \equiv 0 \pmod{16} \] 5. Solve for $ a $: \[ a \equiv -9 \pmod{16} \] Since $ a $ is a base-17 digit, it must be a non-negative integer less than 17. We convert $-9$ to a positive equivalent modulo 16: \[ -9 \equiv 16 - 9 \equiv 7 \pmod{16} \] Thus, $ a = 7 $. The final answer is $ \boxed{7} $.	7	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Andrew writes down all of the prime numbers less than $50$. How many times does he write the digit $2$?	1. First, list all the prime numbers less than $50$. These are: \[ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 \] 2. Identify the prime numbers that contain the digit $2$: - The number $2$ itself contains the digit $2$. - The number $23$ contains the digit $2$. - The number $29$ contains the digit $2$. 3. Count the occurrences of the digit $2$ in these numbers: - The digit $2$ appears once in the number $2$. - The digit $2$ appears once in the number $23$. - The digit $2$ appears once in the number $29$. 4. Sum the occurrences: \[ 1 + 1 + 1 = 3 \] Therefore, the digit $2$ appears $3$ times in the list of prime numbers less than $50$. The final answer is $\boxed{3}$.	3	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
Given that $2x + 5 - 3x + 7 = 8$, what is the value of $x$? $\textbf{(A) }{-}4\qquad\textbf{(B) }{-}2\qquad\textbf{(C) }0\qquad\textbf{(D) }2\qquad\textbf{(E) }4$	1. Start with the given equation: \[ 2x + 5 - 3x + 7 = 8 \] 2. Combine like terms on the left-hand side: \[ (2x - 3x) + (5 + 7) = 8 \] \[ -x + 12 = 8 \] 3. Isolate the variable $x$ by subtracting 12 from both sides: \[ -x + 12 - 12 = 8 - 12 \] \[ -x = -4 \] 4. Solve for $x$ by multiplying both sides by -1: \[ x = 4 \] The final answer is $\boxed{4}$	4	Algebra	MCQ	Yes	Yes	aops_forum	false
An integer $N$ which satisfies exactly three of the four following conditions is called [i]two-good[/i]. $~$ [center] (I) $N$ is divisible by $2$ (II) $N$ is divisible by $4$ (III) $N$ is divisible by $8$ (IV) $N$ is divisible by $16$ [/center]$~$ How many integers between $1$ and $100$, inclusive, are [i]two-good[/i]? $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$	To solve the problem, we need to identify integers $ N $ between 1 and 100 that satisfy exactly three of the four given conditions: 1. $ N $ is divisible by 2. 2. $ N $ is divisible by 4. 3. $ N $ is divisible by 8. 4. $ N $ is divisible by 16. We will analyze each condition and determine which integers satisfy exactly three of these conditions. 1. Condition Analysis: - If $ N $ is divisible by 16, it is also divisible by 8, 4, and 2. Therefore, it satisfies all four conditions, which does not meet the requirement of being two-good. - If $ N $ is divisible by 8 but not by 16, it satisfies conditions (I), (II), and (III) but not (IV). This meets the requirement of being two-good. - If $ N $ is divisible by 4 but not by 8, it satisfies conditions (I) and (II) but not (III) and (IV). This does not meet the requirement of being two-good. - If $ N $ is divisible by 2 but not by 4, it satisfies condition (I) but not (II), (III), and (IV). This does not meet the requirement of being two-good. 2. Finding Two-Good Numbers: - Numbers divisible by 8 but not by 16: These numbers are of the form $ 8k $ where $ k $ is an odd integer. - We need to find such numbers between 1 and 100. 3. Calculation: - The smallest number divisible by 8 is 8. - The largest number divisible by 8 within 100 is 96. - The sequence of numbers divisible by 8 is: $ 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96 $. - We need to exclude numbers that are divisible by 16: $ 16, 32, 48, 64, 80, 96 $. 4. Remaining Numbers: - The numbers that are divisible by 8 but not by 16 are: $ 8, 24, 40, 56, 72, 88 $. 5. Count: - There are 6 such numbers. The final answer is $\boxed{6}$.	6	Logic and Puzzles	MCQ	Yes	Yes	aops_forum	false
Let $x$ be a number such that $10000x+2=4$. What is the value of $5000x+1$? $\textbf{(A) }{-}1\qquad\textbf{(B) }0\qquad\textbf{(C) }1\qquad\textbf{(D) }2\qquad\textbf{(E) }3$	1. Start with the given equation: \[ 10000x + 2 = 4 \] 2. Subtract 2 from both sides to isolate the term with $x$: \[ 10000x + 2 - 2 = 4 - 2 \] \[ 10000x = 2 \] 3. Divide both sides by 10000 to solve for $x$: \[ x = \frac{2}{10000} \] \[ x = \frac{1}{5000} \] 4. Now, we need to find the value of $5000x + 1$. Substitute $x = \frac{1}{5000}$ into the expression: \[ 5000x + 1 = 5000 \left(\frac{1}{5000}\right) + 1 \] \[ 5000x + 1 = 1 + 1 \] \[ 5000x + 1 = 2 \] Conclusion: \[ \boxed{2} \]	2	Algebra	MCQ	Yes	Yes	aops_forum	false
Real numbers $w$, $x$, $y$, and $z$ satisfy $w+x+y = 3$, $x+y+z = 4,$ and $w+x+y+z = 5$. What is the value of $x+y$? $\textbf{(A) }-\frac{1}{2}\qquad\textbf{(B) }1\qquad\textbf{(C) }\frac{3}{2}\qquad\textbf{(D) }2\qquad\textbf{(E) }3$	1. We start with the given equations: \[ w + x + y = 3 \] \[ x + y + z = 4 \] \[ w + x + y + z = 5 \] 2. Subtract the first equation from the third equation: \[ (w + x + y + z) - (w + x + y) = 5 - 3 \] Simplifying, we get: \[ z = 2 \] 3. Substitute $z = 2$ into the second equation: \[ x + y + 2 = 4 \] Simplifying, we get: \[ x + y = 2 \] Thus, the value of $x + y$ is $\boxed{2}$.	2	Algebra	MCQ	Yes	Yes	aops_forum	false
What is the maximum possible value of $5-\|6x-80\|$ over all integers $x$? $\textbf{(A) }{-}1\qquad\textbf{(B) }0\qquad\textbf{(C) }1\qquad\textbf{(D) }3\qquad\textbf{(E) }5$	1. We start with the expression $5 - \|6x - 80\|$. To maximize this expression, we need to minimize the absolute value term $\|6x - 80\|$. 2. The absolute value $\|6x - 80\|$ is minimized when $6x - 80$ is as close to 0 as possible. This occurs when $6x = 80$. 3. Solving for $x$, we get: \[ x = \frac{80}{6} = \frac{40}{3} \approx 13.3333 \] Since $x$ must be an integer, we consider the closest integers to $\frac{40}{3}$, which are $x = 13$ and $x = 14$. 4. Calculate $\|6x - 80\|$ for $x = 13$: \[ 6 \cdot 13 - 80 = 78 - 80 = -2 \quad \Rightarrow \quad \|6 \cdot 13 - 80\| = \| -2 \| = 2 \] 5. Calculate $\|6x - 80\|$ for $x = 14$: \[ 6 \cdot 14 - 80 = 84 - 80 = 4 \quad \Rightarrow \quad \|6 \cdot 14 - 80\| = \| 4 \| = 4 \] 6. Since $\|6 \cdot 13 - 80\| = 2$ is smaller than $\|6 \cdot 14 - 80\| = 4$, the minimum value of $\|6x - 80\|$ is 2, which occurs at $x = 13$. 7. Substitute this value back into the original expression: \[ 5 - \|6 \cdot 13 - 80\| = 5 - 2 = 3 \] The final answer is $\boxed{3}$.	3	Inequalities	MCQ	Yes	Yes	aops_forum	false
Two truth tellers and two liars are positioned in a line, where every person is distinguishable. How many ways are there to position these four people such that everyone claims that all people directly adjacent to them are liars? $\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }12\qquad\textbf{(E) }16$	1. Let the truth tellers be denoted as $ T_1 $ and $ T_2 $, and the liars be denoted as $ L_1 $ and $ L_2 $. 2. Since a liar claims that all people directly adjacent to them are liars, the people directly adjacent to a liar must actually be truth-tellers. Therefore, no liar can be next to another liar, and no truth-teller can be next to another truth-teller. 3. This implies that the arrangement must alternate between truth-tellers and liars. The possible alternating sequences are $ TLTL $ and $ LTLT $. 4. For the sequence $ TLTL $: - We can choose any of the 2 truth-tellers for the first $ T $ position, and any of the remaining 1 truth-teller for the second $ T $ position. This gives us $ 2! = 2 $ ways to arrange the truth-tellers. - Similarly, we can choose any of the 2 liars for the first $ L $ position, and any of the remaining 1 liar for the second $ L $ position. This gives us $ 2! = 2 $ ways to arrange the liars. - Therefore, the total number of ways to arrange $ TLTL $ is $ 2 \times 2 = 4 $. 5. For the sequence $ LTLT $: - We can choose any of the 2 liars for the first $ L $ position, and any of the remaining 1 liar for the second $ L $ position. This gives us $ 2! = 2 $ ways to arrange the liars. - Similarly, we can choose any of the 2 truth-tellers for the first $ T $ position, and any of the remaining 1 truth-teller for the second $ T $ position. This gives us $ 2! = 2 $ ways to arrange the truth-tellers. - Therefore, the total number of ways to arrange $ LTLT $ is $ 2 \times 2 = 4 $. 6. Adding the number of ways for both sequences, we get $ 4 + 4 = 8 $. The final answer is $\boxed{8}$.	8	Logic and Puzzles	MCQ	Yes	Yes	aops_forum	false
An ant climbs either two inches or three inches each day. In how many ways can the ant climb twelve inches, if the order of its climbing sequence matters? $\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }14$	To determine the number of ways the ant can climb twelve inches, we need to consider the different combinations of 2-inch and 3-inch climbs that sum to 12 inches. We will analyze the problem by considering the number of days the ant climbs and the combinations of 2-inch and 3-inch climbs. 1. Case 1: The ant climbs 12 inches in 4 days. - The only way to achieve this is by climbing 3 inches each day. - Sequence: $(3, 3, 3, 3)$ - Number of ways: $1$ 2. Case 2: The ant climbs 12 inches in 5 days. - The ant must climb 3 inches on exactly 2 of the days and 2 inches on the remaining 3 days. - We need to choose 2 days out of 5 for the 3-inch climbs. - Number of ways: $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = 10$ 3. Case 3: The ant climbs 12 inches in 6 days. - The only way to achieve this is by climbing 2 inches each day. - Sequence: $(2, 2, 2, 2, 2, 2)$ - Number of ways: $1$ Adding up all the possible ways from each case, we get: \[ 1 + 10 + 1 = 12 \] Thus, the total number of ways the ant can climb twelve inches is $\boxed{12}$.	12	Combinatorics	MCQ	Yes	Yes	aops_forum	false
[u]Set 1[/u] [b]B1.[/b] Evaluate $2 + 0 - 2 \times 0$. [b]B2.[/b] It takes four painters four hours to paint four houses. How many hours does it take forty painters to paint forty houses? [b]B3.[/b] Let $a$ be the answer to this question. What is $\frac{1}{2-a}$? [u]Set 2[/u] [b]B4.[/b] Every day at Andover is either sunny or rainy. If today is sunny, there is a $60\%$ chance that tomorrow is sunny and a $40\%$ chance that tomorrow is rainy. On the other hand, if today is rainy, there is a $60\%$ chance that tomorrow is rainy and a $40\%$ chance that tomorrow is sunny. Given that today is sunny, the probability that the day after tomorrow is sunny can be expressed as n%, where n is a positive integer. What is $n$? [b]B5.[/b] In the diagram below, what is the value of $\angle DD'Y$ in degrees? [img]https://cdn.artofproblemsolving.com/attachments/0/8/6c966b13c840fa1885948d0e4ad598f36bee9d.png[/img] [b]B6.[/b] Christina, Jeremy, Will, and Nathan are standing in a line. In how many ways can they be arranged such that Christina is to the left of Will and Jeremy is to the left of Nathan? Note: Christina does not have to be next to Will and Jeremy does not have to be next to Nathan. For example, arranging them as Christina, Jeremy, Will, Nathan would be valid. [u]Set 3[/u] [b]B7.[/b] Let $P$ be a point on side $AB$ of square $ABCD$ with side length $8$ such that $PA = 3$. Let $Q$ be a point on side $AD$ such that $P Q \perp P C$. The area of quadrilateral $PQDB$ can be expressed in the form $m/n$ for relatively prime positive integers $m$ and $n$. Compute $m + n$. [b]B8.[/b] Jessica and Jeffrey each pick a number uniformly at random from the set $\{1, 2, 3, 4, 5\}$ (they could pick the same number). If Jessica’s number is $x$ and Jeffrey’s number is $y$, the probability that $x^y$ has a units digit of $1$ can be expressed as $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [b]B9.[/b] For two points $(x_1, y_1)$ and $(x_2, y_2)$ in the plane, we define the taxicab distance between them as $\|x_1 - x_2\| + \|y_1 - y_2\|$. For example, the taxicab distance between $(-1, 2)$ and $(3,\sqrt2)$ is $6-\sqrt2$. What is the largest number of points Nathan can find in the plane such that the taxicab distance between any two of the points is the same? [u]Set 4[/u] [b]B10.[/b] Will wants to insert some × symbols between the following numbers: $$1\,\,\,2\,\,\,3\,\,\,4\,\,\,6$$ to see what kinds of answers he can get. For example, here is one way he can insert $\times$ symbols: $$1 \times 23 \times 4 \times 6 = 552.$$ Will discovers that he can obtain the number $276$. What is the sum of the numbers that he multiplied together to get $276$? [b]B11.[/b] Let $ABCD$ be a parallelogram with $AB = 5$, $BC = 3$, and $\angle BAD = 60^o$ . Let the angle bisector of $\angle ADC$ meet $AC$ at $E$ and $AB$ at $F$. The length $EF$ can be expressed as $m/n$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$? [b]B12.[/b] Find the sum of all positive integers $n$ such that $\lfloor \sqrt{n^2 - 2n + 19} \rfloor = n$. Note: $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. [u]Set 5[/u] [b]B13.[/b] This year, February $29$ fell on a Saturday. What is the next year in which February $29$ will be a Saturday? [b]B14.[/b] Let $f(x) = \frac{1}{x} - 1$. Evaluate $$f\left( \frac{1}{2020}\right) \times f\left( \frac{2}{2020}\right) \times f\left( \frac{3}{2020}\right) \times \times ... \times f\left( \frac{2019}{2020}\right) .$$ [b]B15.[/b] Square $WXYZ$ is inscribed in square $ABCD$ with side length $1$ such that $W$ is on $AB$, $X$ is on $BC$, $Y$ is on $CD$, and $Z$ is on $DA$. Line $W Y$ hits $AD$ and $BC$ at points $P$ and $R$ respectively, and line $XZ$ hits $AB$ and $CD$ at points $Q$ and $S$ respectively. If the area of $WXYZ$ is $\frac{13}{18}$ , then the area of $PQRS$ can be expressed as $m/n$ for relatively prime positive integers $m$ and $n$. What is $m + n$? PS. You had better use hide for answers. Last sets have been posted [url=https://artofproblemsolving.com/community/c4h2777424p24371574]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. Evaluate $2 + 0 - 2 \times 0$. \[ 2 + 0 - 2 \times 0 = 2 + 0 - 0 = 2 \] The final answer is $\boxed{2}$.	2	Other	math-word-problem	Yes	Yes	aops_forum	false
Suppose $(a,b)$ is an ordered pair of integers such that the three numbers $a$, $b$, and $ab$ form an arithmetic progression, in that order. Find the sum of all possible values of $a$. [i]Proposed by Nathan Xiong[/i]	1. Since $a, b, ab$ are in arithmetic progression, the common difference between consecutive terms must be the same. Therefore, we can write the relationship as: \[ b - a = ab - b \] Simplifying this equation, we get: \[ b - a = ab - b \implies 2b = a + ab \] 2. Rearrange the equation to isolate terms involving $a$ and $b$: \[ 2b = a + ab \implies a(b + 1) = 2b \] 3. To solve for $a$, we can use Simon's Favorite Factoring Trick. Rewrite the equation as: \[ a(b + 1) - 2(b + 1) = -2 \implies (a - 2)(b + 1) = -2 \] 4. Now, we need to find all integer pairs $(a - 2, b + 1)$ that satisfy $(a - 2)(b + 1) = -2$. The factor pairs of $-2$ are: \[ (-1, 2), (1, -2), (-2, 1), (2, -1) \] 5. For each factor pair, solve for $a$ and $b$: - If $a - 2 = -1$ and $b + 1 = 2$: \[ a = -1 + 2 = 1, \quad b = 2 - 1 = 1 \] - If $a - 2 = 1$ and $b + 1 = -2$: \[ a = 1 + 2 = 3, \quad b = -2 - 1 = -3 \] - If $a - 2 = -2$ and $b + 1 = 1$: \[ a = -2 + 2 = 0, \quad b = 1 - 1 = 0 \] - If $a - 2 = 2$ and $b + 1 = -1$: \[ a = 2 + 2 = 4, \quad b = -1 - 1 = -2 \] 6. The possible values of $a$ are $0, 1, 3, 4$. Summing these values: \[ 0 + 1 + 3 + 4 = 8 \] The final answer is $\boxed{8}$.	8	Number Theory	math-word-problem	Yes	Yes	aops_forum	false
If $x$, $y$, $z$ are nonnegative integers satisfying the equation below, then compute $x+y+z$. \[\left(\frac{16}{3}\right)^x\times \left(\frac{27}{25}\right)^y\times \left(\frac{5}{4}\right)^z=256.\] [i]Proposed by Jeffrey Shi[/i]	1. Start with the given equation: \[ \left(\frac{16}{3}\right)^x \times \left(\frac{27}{25}\right)^y \times \left(\frac{5}{4}\right)^z = 256 \] 2. Rewrite each fraction in terms of prime factors: \[ \frac{16}{3} = \frac{2^4}{3}, \quad \frac{27}{25} = \frac{3^3}{5^2}, \quad \frac{5}{4} = \frac{5}{2^2} \] 3. Substitute these into the equation: \[ \left(\frac{2^4}{3}\right)^x \times \left(\frac{3^3}{5^2}\right)^y \times \left(\frac{5}{2^2}\right)^z = 256 \] 4. Simplify the equation: \[ \frac{2^{4x}}{3^x} \times \frac{3^{3y}}{5^{2y}} \times \frac{5^z}{2^{2z}} = 256 \] 5. Combine the terms: \[ \frac{2^{4x} \times 3^{3y} \times 5^z}{3^x \times 5^{2y} \times 2^{2z}} = 256 \] 6. Simplify the exponents: \[ 2^{4x-2z} \times 3^{3y-x} \times 5^{z-2y} = 2^8 \times 3^0 \times 5^0 \] 7. Equate the exponents of the corresponding prime factors: \[ 4x - 2z = 8 \quad (1) \] \[ 3y - x = 0 \quad (2) \] \[ z - 2y = 0 \quad (3) \] 8. Solve the system of equations: - From equation (2): \[ x = 3y \] - From equation (3): \[ z = 2y \] - Substitute $ x = 3y $ and $ z = 2y $ into equation (1): \[ 4(3y) - 2(2y) = 8 \] \[ 12y - 4y = 8 \] \[ 8y = 8 \] \[ y = 1 \] - Using $ y = 1 $: \[ x = 3y = 3 \times 1 = 3 \] \[ z = 2y = 2 \times 1 = 2 \] 9. Therefore, the values of $ x, y, z $ are: \[ (x, y, z) = (3, 1, 2) \] 10. Compute $ x + y + z $: \[ x + y + z = 3 + 1 + 2 = 6 \] The final answer is $\boxed{6}$	6	Algebra	math-word-problem	Yes	Yes	aops_forum	false
Four students Alice, Bob, Charlie, and Diana want to arrange themselves in a line such that Alice is at either end of the line, i.e., she is not in between two students. In how many ways can the students do this? [i]Proposed by Nathan Xiong[/i]	1. Determine the possible positions for Alice: - Alice can either be at the first position or the last position in the line. Therefore, there are 2 choices for Alice's position. 2. Arrange the remaining students: - Once Alice's position is fixed, we need to arrange the remaining three students (Bob, Charlie, and Diana) in the remaining three positions. - The number of ways to arrange 3 students is given by the factorial of 3, which is $3!$. 3. Calculate the total number of arrangements: - Since there are 2 choices for Alice's position and $3!$ ways to arrange the remaining students, the total number of arrangements is: \[ 2 \times 3! = 2 \times 6 = 12 \] Conclusion: \[ \boxed{12} \]	12	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
Andover has a special weather forecast this week. On Monday, there is a $\frac{1}{2}$ chance of rain. On Tuesday, there is a $\frac{1}{3}$ chance of rain. This pattern continues all the way to Sunday, when there is a $\frac{1}{8}$ chance of rain. The probability that it doesn't rain in Andover all week can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Compute $m+n$. [i]Proposed by Nathan Xiong[/i]	1. The probability that it doesn't rain on a given day is the complement of the probability that it rains. For Monday, the probability that it doesn't rain is: \[ 1 - \frac{1}{2} = \frac{1}{2} \] 2. For Tuesday, the probability that it doesn't rain is: \[ 1 - \frac{1}{3} = \frac{2}{3} \] 3. This pattern continues for each day of the week. Therefore, the probabilities for each day are: \[ \begin{aligned} &\text{Monday: } \frac{1}{2}, \\ &\text{Tuesday: } \frac{2}{3}, \\ &\text{Wednesday: } \frac{3}{4}, \\ &\text{Thursday: } \frac{4}{5}, \\ &\text{Friday: } \frac{5}{6}, \\ &\text{Saturday: } \frac{6}{7}, \\ &\text{Sunday: } \frac{7}{8}. \end{aligned} \] 4. The probability that it doesn't rain all week is the product of the probabilities that it doesn't rain each day: \[ \left(1 - \frac{1}{2}\right) \cdot \left(1 - \frac{1}{3}\right) \cdot \left(1 - \frac{1}{4}\right) \cdot \left(1 - \frac{1}{5}\right) \cdot \left(1 - \frac{1}{6}\right) \cdot \left(1 - \frac{1}{7}\right) \cdot \left(1 - \frac{1}{8}\right) \] 5. Simplifying each term, we get: \[ \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7} \cdot \frac{7}{8} \] 6. Notice that the numerator and denominator have a telescoping pattern: \[ \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} \] 7. The terms in the numerator and denominator cancel out, leaving: \[ \frac{1}{8} \] 8. The probability that it doesn't rain all week is therefore: \[ \frac{1}{8} \] 9. The fraction $\frac{1}{8}$ is already in simplest form, with $m = 1$ and $n = 8$. Therefore, $m + n = 1 + 8 = 9$. The final answer is $ \boxed{ 9 } $.	9	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
Andy and Harry are trying to make an O for the MOAA logo. Andy starts with a circular piece of leather with radius 3 feet and cuts out a circle with radius 2 feet from the middle. Harry starts with a square piece of leather with side length 3 feet and cuts out a square with side length 2 feet from the middle. In square feet, what is the positive difference in area between Andy and Harry's final product to the nearest integer? [i]Proposed by Andy Xu[/i]	1. Calculate the area of Andy's final product: - Andy starts with a circular piece of leather with radius 3 feet. - The area of the initial circle is given by: \[ A_{\text{initial}} = \pi \times (3)^2 = 9\pi \text{ square feet} \] - Andy cuts out a smaller circle with radius 2 feet from the middle. - The area of the cut-out circle is: \[ A_{\text{cut-out}} = \pi \times (2)^2 = 4\pi \text{ square feet} \] - The area of Andy's final product (the ring) is: \[ A_{\text{Andy}} = 9\pi - 4\pi = 5\pi \text{ square feet} \] 2. Calculate the area of Harry's final product: - Harry starts with a square piece of leather with side length 3 feet. - The area of the initial square is: \[ A_{\text{initial}} = 3 \times 3 = 9 \text{ square feet} \] - Harry cuts out a smaller square with side length 2 feet from the middle. - The area of the cut-out square is: \[ A_{\text{cut-out}} = 2 \times 2 = 4 \text{ square feet} \] - The area of Harry's final product (the frame) is: \[ A_{\text{Harry}} = 9 - 4 = 5 \text{ square feet} \] 3. Calculate the positive difference in area between Andy's and Harry's final products: - The difference in area is: \[ \Delta A = 5\pi - 5 \] - Using the approximation $\pi \approx 3.14$: \[ \Delta A \approx 5 \times 3.14 - 5 = 15.7 - 5 = 10.7 \] - Rounding to the nearest integer: \[ \Delta A \approx 11 \] The final answer is $\boxed{11}$.	11	Geometry	math-word-problem	Yes	Yes	aops_forum	false
Mr. Stein is ordering a two-course dessert at a restaurant. For each course, he can choose to eat pie, cake, rødgrød, and crème brûlée, but he doesn't want to have the same dessert twice. In how many ways can Mr. Stein order his meal? (Order matters.)	1. Mr. Stein has 4 options for the first course: pie, cake, rødgrød, and crème brûlée. 2. Since he doesn't want to have the same dessert twice, he will have 3 remaining options for the second course after choosing the first dessert. 3. Therefore, the number of ways to choose and order the two desserts is calculated by multiplying the number of choices for each course: \[ 4 \text{ (choices for the first course)} \times 3 \text{ (choices for the second course)} = 12 \] The final answer is $\boxed{12}$.	12	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
How many lattice points are exactly twice as close to $(0,0)$ as they are to $(15,0)$? (A lattice point is a point $(a,b)$ such that both $a$ and $b$ are integers.)	1. Set up the distance equations: - The distance from a point $(x, y)$ to $(0, 0)$ is given by: \[ d_1 = \sqrt{x^2 + y^2} \] - The distance from the point $(x, y)$ to $(15, 0)$ is given by: \[ d_2 = \sqrt{(x - 15)^2 + y^2} \] - According to the problem, the point $(x, y)$ is twice as close to $(0, 0)$ as it is to $(15, 0)$. Therefore, we have: \[ d_1 = \frac{1}{2} d_2 \] 2. Square both sides to eliminate the square roots: \[ x^2 + y^2 = \left(\frac{1}{2} \sqrt{(x - 15)^2 + y^2}\right)^2 \] \[ x^2 + y^2 = \frac{1}{4} ((x - 15)^2 + y^2) \] 3. Multiply both sides by 4 to clear the fraction: \[ 4(x^2 + y^2) = (x - 15)^2 + y^2 \] 4. Expand and simplify the equation: \[ 4x^2 + 4y^2 = x^2 - 30x + 225 + y^2 \] \[ 4x^2 + 4y^2 = x^2 + y^2 - 30x + 225 \] 5. Combine like terms: \[ 4x^2 + 4y^2 - x^2 - y^2 = -30x + 225 \] \[ 3x^2 + 3y^2 = -30x + 225 \] 6. Divide the entire equation by 3: \[ x^2 + y^2 = -10x + 75 \] 7. Rearrange the equation to form a circle equation: \[ x^2 + 10x + y^2 = 75 \] \[ (x + 5)^2 - 25 + y^2 = 75 \] \[ (x + 5)^2 + y^2 = 100 \] 8. Identify the circle's center and radius: - The equation $(x + 5)^2 + y^2 = 100$ represents a circle centered at $(-5, 0)$ with a radius of 10. 9. Find the lattice points on the circle: - We need to find integer solutions $(x, y)$ such that $(x + 5)^2 + y^2 = 100$. 10. List the possible integer pairs $(x + 5, y)$ that satisfy the equation: - $(10, 0)$, $(-10, 0)$, $(0, 10)$, $(0, -10)$, $(6, 8)$, $(6, -8)$, $(-6, 8)$, $(-6, -8)$, $(8, 6)$, $(8, -6)$, $(-8, 6)$, $(-8, -6)$. 11. Convert back to the original coordinates $(x, y)$: - $(x, y) = (5, 0)$, $(-15, 0)$, $(-5, 10)$, $(-5, -10)$, $(1, 8)$, $(1, -8)$, $(-11, 8)$, $(-11, -8)$, $(3, 6)$, $(3, -6)$, $(-13, 6)$, $(-13, -6)$. 12. Count the number of lattice points: - There are 12 lattice points that satisfy the condition. The final answer is $\boxed{12}$	12	Geometry	math-word-problem	Yes	Yes	aops_forum	false
Three real numbers $a$, $b$, and $c$ between $0$ and $1$ are chosen independently and at random. What is the probability that $a + 2b + 3c > 5$?	1. We start by considering the unit cube in the 3-dimensional space where each of the coordinates $a$, $b$, and $c$ ranges from 0 to 1. This cube represents all possible combinations of $a$, $b$, and $c$. 2. We need to find the region within this cube where the inequality $a + 2b + 3c > 5$ holds. This inequality represents a plane in the 3-dimensional space. 3. To understand the region better, we first find the intercepts of the plane $a + 2b + 3c = 5$ with the coordinate axes: - When $b = 0$ and $c = 0$, $a = 5$. However, since $a$ is constrained to be between 0 and 1, this intercept is outside the unit cube. - When $a = 0$ and $c = 0$, $2b = 5$ or $b = \frac{5}{2}$. This is also outside the unit cube. - When $a = 0$ and $b = 0$, $3c = 5$ or $c = \frac{5}{3}$. This is also outside the unit cube. 4. Since all intercepts are outside the unit cube, the plane $a + 2b + 3c = 5$ does not intersect the unit cube. Therefore, the region where $a + 2b + 3c > 5$ lies entirely outside the unit cube. 5. As a result, there are no points $(a, b, c)$ within the unit cube that satisfy the inequality $a + 2b + 3c > 5$. 6. Therefore, the volume of the region within the unit cube where $a + 2b + 3c > 5$ is zero. The final answer is $\boxed{0}$.	0	Inequalities	math-word-problem	Yes	Yes	aops_forum	false
[u]Set 1[/u] [b]p1.[/b] Arnold is currently stationed at $(0, 0)$. He wants to buy some milk at $(3, 0)$, and also some cookies at $(0, 4)$, and then return back home at $(0, 0)$. If Arnold is very lazy and wants to minimize his walking, what is the length of the shortest path he can take? [b]p2.[/b] Dilhan selects $1$ shirt out of $3$ choices, $1$ pair of pants out of $4$ choices, and $2$ socks out of $6$ differently-colored socks. How many outfits can Dilhan select? All socks can be worn on both feet, and outfits where the only difference is that the left sock and right sock are switched are considered the same. [b]p3.[/b] What is the sum of the first $100$ odd positive integers? [b]p4.[/b] Find the sum of all the distinct prime factors of $1591$. [b]p5.[/b] Let set $S = \{1, 2, 3, 4, 5, 6\}$. From $S$, four numbers are selected, with replacement. These numbers are assembled to create a $4$-digit number. How many such $4$-digit numbers are multiples of $3$? [u]Set 2[/u] [b]p6.[/b] What is the area of a triangle with vertices at $(0, 0)$, $(7, 2)$, and $(4, 4)$? [b]p7.[/b] Call a number $n$ “warm” if $n - 1$, $n$, and $n + 1$ are all composite. Call a number $m$ “fuzzy” if $m$ may be expressed as the sum of $3$ consecutive positive integers. How many numbers less than or equal to $30$ are warm and fuzzy? [b]p8.[/b] Consider a square and hexagon of equal area. What is the square of the ratio of the side length of the hexagon to the side length of the square? [b]p9.[/b] If $x^2 + y^2 = 361$, $xy = -40$, and $x - y$ is positive, what is $x - y$? [b]p10.[/b] Each face of a cube is to be painted red, orange, yellow, green, blue, or violet, and each color must be used exactly once. Assuming rotations are indistinguishable, how many ways are there to paint the cube? [u]Set 3[/u] [b]p11.[/b] Let $D$ be the midpoint of side $BC$ of triangle $ABC$. Let $P$ be any point on segment $AD$. If $M$ is the maximum possible value of $\frac{[PAB]}{[PAC]}$ and $m$ is the minimum possible value, what is $M - m$? Note: $[PQR]$ denotes the area of triangle $PQR$. [b]p12.[/b] If the product of the positive divisors of the positive integer $n$ is $n^6$, find the sum of the $3$ smallest possible values of $n$. [b]p13.[/b] Find the product of the magnitudes of the complex roots of the equation $(x - 4)^4 +(x - 2)^4 + 14 = 0$. [b]p14.[/b] If $xy - 20x - 16y = 2016$ and $x$ and $y$ are both positive integers, what is the least possible value of $\max (x, y)$? [b]p15.[/b] A peasant is trying to escape from Chyornarus, ruled by the Tsar and his mystical faith healer. The peasant starts at $(0, 0)$ on a $6 \times 6$ unit grid, the Tsar’s palace is at $(3, 3)$, the healer is at $(2, 1)$, and the escape is at $(6, 6)$. If the peasant crosses the Tsar’s palace or the mystical faith healer, he is executed and fails to escape. The peasant’s path can only consist of moves upward and rightward along the gridlines. How many valid paths allow the peasant to escape? PS. You should use hide for answers. Rest sets have been posted [url=https://artofproblemsolving.com/community/c3h2784259p24464954]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	1. To find the shortest path Arnold can take, we need to consider the distances between the points and the possible paths. The points are $(0,0)$ (home), $(3,0)$ (milk), and $(0,4)$ (cookies). - Distance from $(0,0)$ to $(3,0)$: \[ \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9} = 3 \] - Distance from $(3,0)$ to $(0,4)$: \[ \sqrt{(0-3)^2 + (4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] - Distance from $(0,4)$ to $(0,0)$: \[ \sqrt{(0-0)^2 + (0-4)^2} = \sqrt{16} = 4 \] Adding these distances together, we get: \[ 3 + 5 + 4 = 12 \] The final answer is $\boxed{12}$	12	Geometry	math-word-problem	Yes	Yes	aops_forum	false
[u]Set 4[/u] [b]p16.[/b] Albert, Beatrice, Corey, and Dora are playing a card game with two decks of cards numbered $1-50$ each. Albert, Beatrice, and Corey draw cards from the same deck without replacement, but Dora draws from the other deck. What is the probability that the value of Corey’s card is the highest value or is tied for the highest value of all $4$ drawn cards? [b]p17.[/b] Suppose that $s$ is the sum of all positive values of $x$ that satisfy $2016\{x\} = x+[x]$. Find $\{s\}$. (Note: $[x]$ denotes the greatest integer less than or equal to $x$ and $\{x\}$ denotes $x - [x]$.) [b]p18.[/b] Let $ABC$ be a triangle such that $AB = 41$, $BC = 52$, and $CA = 15$. Let H be the intersection of the $B$ altitude and $C$ altitude. Furthermore let $P$ be a point on $AH$. Both $P$ and $H$ are reflected over $BC$ to form $P'$ and $H'$ . If the area of triangle $P'H'C$ is $60$, compute $PH$. [b]p19.[/b] A random integer $n$ is chosen between $1$ and $30$, inclusive. Then, a random positive divisor of $n, k$, is chosen. What is the probability that $k^2 > n$? [b]p20.[/b] What are the last two digits of the value $3^{361}$? [u]Set 5[/u] [b]p21.[/b] Let $f(n)$ denote the number of ways a $3 \times n$ board can be completely tiled with $1 \times 3$ and $1 \times 4$ tiles, without overlap or any tiles hanging over the edge. The tiles may be rotated. Find $\sum^9_{i=0} f(i) = f(0) + f(1) + ... + f(8) + f(9)$. By convention, $f(0) = 1$. [b]p22.[/b] Find the sum of all $5$-digit perfect squares whose digits are all distinct and come from the set $\{0, 2, 3, 5, 7, 8\}$. [b]p23.[/b] Mary is flipping a fair coin. On average, how many flips would it take for Mary to get $4$ heads and $2$ tails? [b]p24.[/b] A cylinder is formed by taking the unit circle on the $xy$-plane and extruding it to positive infinity. A plane with equation $z = 1 - x$ truncates the cylinder. As a result, there are three surfaces: a surface along the lateral side of the cylinder, an ellipse formed by the intersection of the plane and the cylinder, and the unit circle. What is the total surface area of the ellipse formed and the lateral surface? (The area of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is $\pi ab$.) [b]p25.[/b] Let the Blair numbers be defined as follows: $B_0 = 5$, $B_1 = 1$, and $B_n = B_{n-1} + B_{n-2}$ for all $n \ge 2$. Evaluate $$\sum_{i=0}^{\infty} \frac{B_i}{51^i}= B_0 +\frac{B_1}{51} +\frac{B_2}{51^2} +\frac{B_3}{51^3} +...$$ [u]Estimation[/u] [b]p26.[/b] Choose an integer between $1$ and $10$, inclusive. Your score will be the number you choose divided by the number of teams that chose your number. [b]p27.[/b] $2016$ blind people each bring a hat to a party and leave their hat in a pile at the front door. As each partier leaves, they take a random hat from the ones remaining in a pile. Estimate the probability that at least $1$ person gets their own hat back. [b]p28.[/b] Estimate how many lattice points lie within the graph of $\|x^3\| + \|y^3\| < 2016$. [b]p29.[/b] Consider all ordered pairs of integers $(x, y)$ with $1 \le x, y \le 2016$. Estimate how many such ordered pairs are relatively prime. [b]p30.[/b] Estimate how many times the letter “e” appears among all Guts Round questions. PS. You should use hide for answers. First sets have been posted [url=https://artofproblemsolving.com/community/c3h2779594p24402189]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].	To find the last two digits of $3^{361}$, we can use Euler's theorem. Euler's theorem states that if $a$ and $n$ are coprime, then $a^{\phi(n)} \equiv 1 \pmod{n}$, where $\phi(n)$ is the Euler's totient function. 1. Calculate $\phi(100)$: \[ \phi(100) = \phi(2^2 \cdot 5^2) = 100 \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{5}\right) = 100 \cdot \frac{1}{2} \cdot \frac{4}{5} = 40 \] 2. Apply Euler's theorem: Since $(3, 100) = 1$, we have: \[ 3^{40} \equiv 1 \pmod{100} \] 3. Express $3^{361}$ in terms of $3^{40}$: \[ 3^{361} = 3^{360 + 1} = (3^{40})^9 \cdot 3^1 \] Using Euler's theorem: \[ (3^{40})^9 \equiv 1^9 \equiv 1 \pmod{100} \] Therefore: \[ 3^{361} \equiv 1 \cdot 3 \equiv 3 \pmod{100} \] Thus, the last two digits of $3^{361}$ are $03$. The final answer is $\boxed{03}$.	03	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
A blind ant is walking on the coordinate plane. It is trying to reach an anthill, placed at all points where both the $x$-coordinate and $y$-coordinate are odd. The ant starts at the origin, and each minute it moves one unit either up, down, to the right, or to the left, each with probability $\frac{1}{4}$. The ant moves $3$ times and doesn't reach an anthill during this time. On average, how many additional moves will the ant need to reach an anthill? (Compute the expected number of additional moves needed.)	1. Understanding the Problem: - The ant starts at the origin $(0,0)$. - It moves one unit in one of the four directions (up, down, left, right) with equal probability $\frac{1}{4}$. - The ant needs to reach a point where both coordinates are odd. - We need to find the expected number of additional moves required after the ant has already moved 3 times and hasn't reached an anthill. 2. Possible Positions After 3 Moves: - After 3 moves, the ant can be at any point $(x, y)$ where $x$ and $y$ are integers. - Since the ant hasn't reached an anthill, both $x$ and $y$ cannot be odd simultaneously. 3. Expected Number of Moves to Reach an Anthill: - We need to calculate the expected number of moves for the ant to reach a point where both coordinates are odd. - Let $E(x, y)$ be the expected number of moves from point $(x, y)$ to reach an anthill. 4. Setting Up the Equations: - If the ant is at $(x, y)$ where both $x$ and $y$ are even or one is even and the other is odd, it needs to move to a point where both coordinates are odd. - The expected number of moves from $(x, y)$ can be expressed as: \[ E(x, y) = 1 + \frac{1}{4} \left( E(x+1, y) + E(x-1, y) + E(x, y+1) + E(x, y-1) \right) \] - This equation accounts for the fact that the ant moves one step and then has an equal probability of moving to one of the four neighboring points. 5. Simplifying the Problem: - Consider the symmetry and periodicity of the problem. The ant's position modulo 2 is crucial. - Define $E_{even, even}$, $E_{even, odd}$, $E_{odd, even}$, and $E_{odd, odd}$ as the expected number of moves from points with respective parities. 6. Solving the System of Equations: - From the symmetry and periodicity, we can set up the following system: \[ E_{even, even} = 1 + \frac{1}{4} \left( E_{odd, even} + E_{odd, even} + E_{even, odd} + E_{even, odd} \right) \] \[ E_{even, odd} = 1 + \frac{1}{4} \left( E_{odd, odd} + E_{odd, odd} + E_{even, even} + E_{even, even} \right) \] \[ E_{odd, even} = 1 + \frac{1}{4} \left( E_{odd, odd} + E_{odd, odd} + E_{even, even} + E_{even, even} \right) \] \[ E_{odd, odd} = 0 \] - Simplifying these equations: \[ E_{even, even} = 1 + \frac{1}{2} E_{odd, even} + \frac{1}{2} E_{even, odd} \] \[ E_{even, odd} = 1 + \frac{1}{2} E_{odd, odd} + \frac{1}{2} E_{even, even} \] \[ E_{odd, even} = 1 + \frac{1}{2} E_{odd, odd} + \frac{1}{2} E_{even, even} \] \[ E_{odd, odd} = 0 \] 7. Solving for $E_{even, even}$: - Substitute $E_{odd, odd} = 0$ into the equations: \[ E_{even, odd} = 1 + \frac{1}{2} E_{even, even} \] \[ E_{odd, even} = 1 + \frac{1}{2} E_{even, even} \] - Substitute these into the equation for $E_{even, even}$: \[ E_{even, even} = 1 + \frac{1}{2} (1 + \frac{1}{2} E_{even, even}) + \frac{1}{2} (1 + \frac{1}{2} E_{even, even}) \] \[ E_{even, even} = 1 + \frac{1}{2} + \frac{1}{4} E_{even, even} + \frac{1}{2} + \frac{1}{4} E_{even, even} \] \[ E_{even, even} = 2 + \frac{1}{2} E_{even, even} \] \[ \frac{1}{2} E_{even, even} = 2 \] \[ E_{even, even} = 4 \] 8. Conclusion: - The expected number of additional moves needed for the ant to reach an anthill is $4$. The final answer is $\boxed{4}$	4	Combinatorics	math-word-problem	Yes	Yes	aops_forum	false
Let $a_n =\sum_{d\|n} \frac{1}{2^{d+ \frac{n}{d}}}$. In other words, $a_n$ is the sum of $\frac{1}{2^{d+ \frac{n}{d}}}$ over all divisors $d$ of $n$. Find $$\frac{\sum_{k=1} ^{\infty}ka_k}{\sum_{k=1}^{\infty} a_k} =\frac{a_1 + 2a_2 + 3a_3 + ....}{a_1 + a_2 + a_3 +....}$$	1. We start by analyzing the given sequence $ a_n = \sum_{d\|n} \frac{1}{2^{d + \frac{n}{d}}} $, where the sum is taken over all divisors $ d $ of $ n $. 2. We need to find the value of the expression: \[ \frac{\sum_{k=1}^\infty k a_k}{\sum_{k=1}^\infty a_k} \] 3. First, we compute the numerator $\sum_{k=1}^\infty k a_k$: \[ \sum_{k=1}^\infty k a_k = \sum_{k=1}^\infty k \sum_{d\|k} \frac{1}{2^{d + \frac{k}{d}}} \] By changing the order of summation, we get: \[ \sum_{k=1}^\infty k a_k = \sum_{k=1}^\infty \sum_{d\|k} \frac{k}{2^{d + \frac{k}{d}}} \] Let $ k = d \cdot l $, then: \[ \sum_{k=1}^\infty k a_k = \sum_{d=1}^\infty \sum_{l=1}^\infty \frac{d \cdot l}{2^{d + l}} \] This can be rewritten as: \[ \sum_{d=1}^\infty \frac{d}{2^d} \sum_{l=1}^\infty \frac{l}{2^l} \] 4. We now compute the inner sums: \[ \sum_{l=1}^\infty \frac{l}{2^l} \] Using the formula for the sum of a geometric series and its derivative, we have: \[ \sum_{l=1}^\infty \frac{l}{2^l} = \frac{\frac{1}{2}}{(1 - \frac{1}{2})^2} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2 \] 5. Similarly, for the outer sum: \[ \sum_{d=1}^\infty \frac{d}{2^d} = 2 \] 6. Therefore, the numerator becomes: \[ \sum_{k=1}^\infty k a_k = 2 \cdot 2 = 4 \] 7. Next, we compute the denominator $\sum_{k=1}^\infty a_k$: \[ \sum_{k=1}^\infty a_k = \sum_{k=1}^\infty \sum_{d\|k} \frac{1}{2^{d + \frac{k}{d}}} \] Again, changing the order of summation: \[ \sum_{k=1}^\infty a_k = \sum_{d=1}^\infty \sum_{l=1}^\infty \frac{1}{2^{d + l}} \] This can be rewritten as: \[ \sum_{d=1}^\infty \frac{1}{2^d} \sum_{l=1}^\infty \frac{1}{2^l} \] 8. We compute the inner sums: \[ \sum_{l=1}^\infty \frac{1}{2^l} = 1 \] 9. Similarly, for the outer sum: \[ \sum_{d=1}^\infty \frac{1}{2^d} = 1 \] 10. Therefore, the denominator becomes: \[ \sum_{k=1}^\infty a_k = 1 \cdot 1 = 1 \] 11. Finally, we find the value of the expression: \[ \frac{\sum_{k=1}^\infty k a_k}{\sum_{k=1}^\infty a_k} = \frac{4}{1} = 4 \] The final answer is $\boxed{4}$	4	Number Theory	math-word-problem	Yes	Yes	aops_forum	false

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.