id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
1,601 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_26 | 2 | If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$ | adding $\log_{10} n$ to both sides: \[\log_{10} m + \log_{10} n=b\] using the logarithm property: $\log_a {b} + \log_a {c}=\log_a{bc}$ \[\log_{10} {mn}=b\] rewriting in exponential notation: \[10^b=mn\] \[m=\boxed{10}\] ~Vndom | 10 |
1,602 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_27 | 1 | A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:
$\textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37... | The car takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{30 \text{ miles }}=4 \text{ hr}$ to get from $A$ to $B$ . Also, it takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{40 \text{ miles }}=3 \text{ hr}$ to get from $B$ to $A$ . Therefore, the average speed is $\dfrac{240\text{ miles }}{7 \text{ hr}}=34\dfrac{2... | 34 |
1,603 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_28 | 1 | Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$ $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:
$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph}... | Let the speed of boy $A$ be $a$ , and the speed of boy $B$ be $b$ . Notice that $A$ travels $4$ miles per hour slower than boy $B$ , so we can replace $b$ with $a+4$
Now let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=rt$ to ... | 8 |
1,604 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_30 | 1 | From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \text{... | Let us represent the number of boys $b$ , and the number of girls $g$
From the first sentence, we get that $2(g-15)=b$
From the second sentence, we get $5(b-45)=g-15$
Expanding both equations and simplifying, we get $2g-30 = b$ and $5b = g+210$
Substituting $b$ for $2g-30$ , we get $5(2g-30)=g+210$ . Solving for $g$ , ... | 40 |
1,605 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_32 | 1 | $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textb... | By the Pythagorean triple $(7,24,25)$ , the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \boxed{8}$ | 8 |
1,606 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_32 | 2 | $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textb... | We can observe that the above setup forms a right angled triangles whose base is 7ft and whose hypotenuse is 25ft taking the height to be x ft.
\[x^2 + 7^2 = 25^2\] \[x^2 = 625 - 49\] \[x^2 = 576\] \[x = 24\]
Since the top of the ladder slipped by 4 ft the new height is $24 - 4 = 20 ft$ . The base of the ladder has mov... | 8 |
1,607 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_33 | 1 | The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:
$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$ | It must be assumed that the pipes have an equal height.
We can represent the amount of water carried per unit time by cross sectional area.
Cross sectional of Pipe with diameter $6 in$ \[\pi r^2 = \pi \cdot 3^2 = 9\pi\]
Cross sectional area of pipe with diameter $1 in$
\[\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4... | 36 |
1,608 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_35 | 1 | In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$ | The inradius is equal to the area divided by semiperimeter. The area is $\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$ . Therefore the inradius is $\boxed{4}$ | 4 |
1,609 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_35 | 2 | In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$ | Since this is a right triangle, we have \[\frac{a+b-c}{2}=\boxed{4}\] | 4 |
1,610 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_36 | 1 | A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?
$\textbf{(A)}\ 125\% \qquad \textbf{(B)}\ 100\% \qquad \textbf{(C)... | Without loss of generality, we can set the list price equal to $100$ . The merchant buys the goods for $100*.75=75$ . Let $x$ be the marked price.
We then use the equation $0.8x-75=25$ to solve for $x$ and get a marked price of $\boxed{125}$ | 125 |
1,611 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_38 | 1 | If the expression $\begin{pmatrix}a & c\\ d & b\end{pmatrix}$ has the value $ab-cd$ for all values of $a, b, c$ and $d$ , then the equation $\begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3$
$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \... | By $\begin{pmatrix}a & c\\ d & b\end{pmatrix}=ab-cd$ , we have $2x^2-x=3$ . Subtracting $3$ from both sides, giving $2x^2-x-3=0$ . This factors to $(2x-3)(x+1)=0$ . Thus, $x=\dfrac{3}{2},-1$ , so the equation is $\boxed{2}$ | 2 |
1,612 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_40 | 1 | The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$ | Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$ , using the L'Hôpital's rule, we have $\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2$ .
Thus, the answer is $\boxed{2}$ | 2 |
1,613 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_40 | 2 | The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$ | The numerator of $\frac {x^2-1}{x-1}$ can be factored as $(x+1)(x-1)$ . The $x-1$ terms in the numerator and denominator cancel, so the expression is equal to $x+1$ so long as $x$ does not equal $1$ . Looking at the function's behavior near 1, we see that as $x$ approaches one, the expression approaches $\boxed{2}$ | 2 |
1,614 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45 | 1 | The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$ | Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be $\binom{100}{2}=4950$ . However this also counts the 100 sides of the polygon, so the actual answer is $4950-100=\boxed{4850}$ | 850 |
1,615 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45 | 2 | The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$ | The formula for the number of diagonals of a polygon with $n$ sides is $n(n-3)/2$ . Taking $n=100$ , we see that the number of diagonals that may be drawn in this polygon is $100(97)/2$ or $\boxed{4850}$ | 850 |
1,616 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_46 | 1 | In triangle $ABC$ $AB=12$ $AC=7$ , and $BC=10$ . If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:
$\textbf{(A)}\ \text{The area is doubled} \qquad\\ \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ \textbf{(D)}\ \text{The m... | If you double sides $AB$ and $AC$ , they become $24$ and $14$ respectively. If $BC$ remains $10$ , then this triangle has area $0$ because ${14} + {10} = {24}$ , so two sides overlap the third side. Therefore the answer is $\boxed{0}$ | 0 |
1,617 | https://artofproblemsolving.com/wiki/index.php/2020_USOMO_Problems/Problem_4 | 1 | Suppose that $(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \le i < j \le 100$ and $|a_ib_j - a_j b_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ... | Let's start off with just $(a_1, b_1), (a_2, b_2)$ and suppose that it satisfies the given condition. We could use $(1, 1), (1, 2)$ for example. We should maximize the number of conditions that the third pair satisfies. We find out that the third pair should equal $(a_1+a_2, b_1+b_2)$
We know this must be true: \[|a_1b... | 197 |
1,618 | https://artofproblemsolving.com/wiki/index.php/2020_USOMO_Problems/Problem_4 | 2 | Suppose that $(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \le i < j \le 100$ and $|a_ib_j - a_j b_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ... | We claim the answer is $197$
Study the points $(0, 0), (a_i, b_i), (a_j, b_j)$ . If we let these be the vertices of a triangle, applying shoelace theorem gives us an area of $\frac{1}{2}|0\times{b_i}+{a_i}\times{b_j}+{b_i}\times{0}-0\times{a_i}-{b_i}\times{a_j}-{b_j}\times{0} = \frac{1}{2}|a_ib_j - a_j b_i| = \frac{1}{... | 197 |
1,619 | https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_4 | 1 | problem_id
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
Name: Text, dtype: object | We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\boxed{6}$ | 6 |
1,620 | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | 1 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to s... | 13 |
1,621 | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | 2 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Outline:
1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \ge 3$
2. If the chosen $n$ is such that $F_n \le n^2$ , then choose the sequence $a_n$ such that $a_k = \sqrt{F_k}$ for $1 \le k \le n$ . It is easy to verify that such a sequence satisfies the condition that the larges... | 13 |
1,622 | https://artofproblemsolving.com/wiki/index.php/2000_USAMO_Problems/Problem_4 | 1 | Find the smallest positive integer $n$ such that if $n$ squares of a $1000 \times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board. | We claim that $n = 1999$ is the smallest such number. For $n \le 1998$ , we can simply color any of the $1998$ squares forming the top row and the left column, but excluding the top left corner square.
We now show that no configuration with no colored right triangles exists for $n = 1999$ . We call a row or column fill... | 999 |
1,623 | https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_1 | 1 | In triangle $ABC$ , angle $A$ is twice angle $B$ , angle $C$ is obtuse , and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter | [asy] import olympiad; pair A, B, C, D, extensionAC; real angleABC; path braceBC; A = (0, 0); B = (2, 0); D = (1, .5); angleABC = atan(.5); //y = 4x/3 and x+2y = 2 (sides AC and BC, respectively) intersect here: C = (6/11, 8/11); braceBC = brace(C, B, .1); label("$\mathsf{A}$", A, W); label("$\mathsf{B}$", B, E);... | 77 |
1,624 | https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_1 | 2 | In triangle $ABC$ , angle $A$ is twice angle $B$ , angle $C$ is obtuse , and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter | In $\triangle ABC$ let $\angle B = \beta, \angle A = 2\beta, \angle C = 180^{\circ} - 3\beta$ . From the law of sines, we have \[\frac{a}{\sin 2\beta} = \frac{b}{\sin \beta} = \frac{c} {\sin (180^{\circ} - 3\beta)} = \frac{c}{\sin 3\beta}\] Thus the ratio \[b : a : c = \sin\beta : \sin 2\beta : \sin 3\beta\] We can sim... | 77 |
1,625 | https://artofproblemsolving.com/wiki/index.php/1986_USAMO_Problems/Problem_3 | 1 | What is the smallest integer $n$ , greater than one, for which the root-mean-square of the first $n$ positive integers is an integer?
$\mathbf{Note.}$ The root-mean-square of $n$ numbers $a_1, a_2, \cdots, a_n$ is defined to be \[\left[\frac{a_1^2 + a_2^2 + \cdots + a_n^2}n\right]^{1/2}\] | Let's first obtain an algebraic expression for the root mean square of the first $n$ integers, which we denote $I_n$ . By repeatedly using the identity $(x+1)^3 = x^3 + 3x^2 + 3x + 1$ , we can write \[1^3 + 3\cdot 1^2 + 3 \cdot 1 + 1 = 2^3,\] \[1^3 + 3 \cdot(1^2 + 2^2) + 3 \cdot (1 + 2) + 1 + 1 = 3^3,\] and \[1^3 + 3\... | 337 |
1,626 | https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_5 | 1 | Let $a_1,a_2,a_3,\cdots$ be a non-decreasing sequence of positive integers. For $m\ge1$ , define $b_m=\min\{n: a_n \ge m\}$ , that is, $b_m$ is the minimum value of $n$ such that $a_n\ge m$ . If $a_{19}=85$ , determine the maximum value of $a_1+a_2+\cdots+a_{19}+b_1+b_2+\cdots+b_{85}$ | We create an array of dots like so: the array shall go out infinitely to the right and downwards, and at the top of the $i$ th column we fill the first $a_i$ cells with one dot each. Then the $19$ th row shall have 85 dots. Now consider the first 19 columns of this array, and consider the first 85 rows. In row $j$ , we... | 700 |
1,627 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | 1 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | Using Vieta's formulas, we have:
\begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*}
From the last of these equations, we see that $cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62$ . Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$ , and so $ac+ad+bc+bd=... | 86 |
1,628 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | 2 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | We start as before: $ab=-32$ and $cd=62$ . We now observe that a and b must be the roots of a quadratic, $x^2+rx-32$ , where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic $x^2+sx+62$
Now
\begin{align*}x^4-18x^3+kx^2+200x-1984 =& (x^2+rx-32)(x^2+sx+... | 86 |
1,629 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | 3 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | Let the roots of the equation be $a,b,c,$ and $d$ . By Vieta's, \begin{align*} a+b+c+d &= 18\\ ab+ac+ad+bc+bd+cd &= k\\ abc+abd+acd+bcd &=-200\\ abcd &=-1984.\\ \end{align*} Since $abcd=-1984$ and $ab=-32$ , then, $cd=62$ . Notice that \[abc + abd + acd + bcd = -200\] can be factored into \[ab(c+d)+cd(a+b)=-32(c+d)+... | 86 |
1,630 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | 4 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | Since two of the roots have product $-32,$ the equation can be factored in the form \[x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + c).\] Expanding, we get \[x^4 - 18x^3 + kx^2 + 200x - 1984 = x^4 + (a + b) x^3 + (ab + c - 32) x^2 + (ac - 32b) x - 32c = 0.\] Matching coefficients, we get
\begin{align*}
... | 86 |
1,631 | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_1 | 1 | In a party with $1982$ people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else? | We induct on $n$ to prove that in a party with $n$ people, there must be at least $(n-3)$ people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people $A, B, C$ , who do not know each other.)
Base case: $n = 4$ is obvious.
Inductive step: Suppose in a party with $... | 979 |
1,632 | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_2 | 1 | Let $S_r=x^r+y^r+z^r$ with $x,y,z$ real. It is known that if $S_1=0$
$(*)$ $\frac{S_{m+n}}{m+n}=\frac{S_m}{m}\frac{S_n}{n}$
for $(m,n)=(2,3),(3,2),(2,5)$ , or $(5,2)$ . Determine all other pairs of integers $(m,n)$ if any, so that $(*)$ holds for all real numbers $x,y,z$ such that $x+y+z=0$ | Claim Both $m,n$ can not be even.
Proof $x+y+z=0$ $\implies x=-(y+z)$
Since $\frac{S_{m+n}}{m+n} = \frac{S_m S_n}{mn}$
by equating cofficient of $y^{m+n}$ on LHS and RHS ,get
$\frac{2}{m+n}=\frac{4}{mn}$
$\implies \frac{m}{2} + \frac {n}{2} = \frac{m\cdot n}{2\cdot2}$
So we have, $\frac{m}{2} \biggm{|} \frac{n}{2}$ an... | 2 |
1,633 | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_4 | 1 | Prove that there exists a positive integer $k$ such that $k\cdot2^n+1$ is composite for every integer $n$ | Indeed, $\boxed{2935363331541925531}$ has the requisite property. | 531 |
1,634 | https://artofproblemsolving.com/wiki/index.php/2021_USAJMO_Problems/Problem_1 | 1 | Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\] | The answer is $\boxed{1}$ , which works.
To show it is necessary, we first get $f(1)=f(1)^2$ , so $f(1)=1$ .
Then, we get $f(2)=f(1^2 + 1^2)=f(1)^2 =1$ | 1 |
1,635 | https://artofproblemsolving.com/wiki/index.php/2020_USOJMO_Problems/Problem_5 | 1 | Suppose that $(a_1,b_1),$ $(a_2,b_2),$ $\dots,$ $(a_{100},b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i,j)$ satisfying $1\leq i<j\leq 100$ and $|a_ib_j-a_jb_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ order... | Call the pair $(i, j)$ good if $1\leq i < j \leq 100$ and $|a_ib_j-a_jb_i|=1$ . Note that we can reorder the pairs $(a_1, b_1), (a_2, b_2), \ldots, (a_{100}, b_{100})$ without changing the number of good pairs. Thus, we can reorder them so that $a_1\leq a_2\leq\ldots\leq a_{100}$ . Furthermore, reorder them so that if ... | 197 |
1,636 | https://artofproblemsolving.com/wiki/index.php/2016_USAJMO_Problems/Problem_4 | 1 | Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,...,N\}$ , one can still find $2016$ distinct numbers among the remaining elements with sum $N$ | Since any $2016$ elements are removed, suppose we remove the integers from $1$ to $2016$ . Then the smallest possible sum of $2016$ of the remaining elements is \[2017+2018+\cdots + 4032 = 1008 \cdot 6049 = 6097392\] so clearly $N\ge 6097392$ . We will show that $N=6097392$ works.
$\vspace{0.2 in}$
$\{1,2\cdots 6097392... | 392 |
1,637 | https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_4 | 2 | problem_id
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
Name: Text, dtype: object | We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\boxed{6}$ | 6 |
1,638 | https://artofproblemsolving.com/wiki/index.php/2013_USAJMO_Problems/Problem_4 | 1 | Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$ $2+2$ $2+1+1$ $1+2+1$ $1+1+2$ , and $1+1+1+1$ . Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd. | First of all, note that $f(n)$ $\sum_{i=0}^{k} f(n-2^{i})$ where $k$ is the largest integer such that $2^k \le n$ . We let $f(0) = 1$ for convenience.
From here, we proceed by induction, with our claim being that the only $n$ such that $f(n)$ is odd are $n$ representable of the form $2^{a} - 1, a \in \mathbb{Z}$
We in... | 47 |
1,639 | https://artofproblemsolving.com/wiki/index.php/2013_USAJMO_Problems/Problem_4 | 2 | Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$ $2+2$ $2+1+1$ $1+2+1$ $1+1+2$ , and $1+1+1+1$ . Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd. | Of course, as with any number theory problem, use actual numbers to start, not variables! By plotting out the first few sums (do it!) and looking for patterns, we observe that $f(n)=\sum_{\textrm{power}=0}^{\textrm{pow}_{\textrm{larg}}} f(n-2^{\textrm{power}})$ , where $\textrm{pow}_{\textrm{larg}}$ represents the larg... | 47 |
1,640 | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | 3 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to s... | 13 |
1,641 | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | 4 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Outline:
1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \ge 3$
2. If the chosen $n$ is such that $F_n \le n^2$ , then choose the sequence $a_n$ such that $a_k = \sqrt{F_k}$ for $1 \le k \le n$ . It is easy to verify that such a sequence satisfies the condition that the larges... | 13 |
1,642 | https://artofproblemsolving.com/wiki/index.php/2012_USAJMO_Problems/Problem_5 | 1 | For distinct positive integers $a$ $b < 2012$ , define $f(a,b)$ to be the number of integers $k$ with $1 \le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012. Let $S$ be the minimum value of $f(a,b)$ , where $a$ and $b$ range over all pairs of distinct positive i... | Let $ak \equiv r_{a} \pmod{2012}$ and $bk \equiv r_{b} \pmod{2012}$ . Notice that this means $a(2012 - k) \equiv 2012 - r_{a} \pmod{2012}$ and $b(2012 - k) \equiv 2012 - r_{b} \pmod{2012}$ . Thus, for every value of $k$ where $r_{a} > r_{b}$ , there is a value of $k$ where $r_{b} > r_{a}$ . Therefore, we merely have to... | 502 |
1,643 | https://artofproblemsolving.com/wiki/index.php/2011_USAJMO_Problems/Problem_1 | 1 | Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square. | We will first take the expression modulo $3$ . We get $2^n+12^n+2011^n \equiv -1^n+1^n \pmod 3$
Lemma 1: All perfect squares are equal to $0$ or $1$ modulo $3$ .
We can prove this by testing the residues modulo $3$ . We have $0^2 \equiv 0 \pmod 3$ $1^2 \equiv 1 \pmod 3$ , and $2^2 \equiv 1 \pmod 3$ , so the lemma is tr... | 1 |
1,644 | https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1 | 1 | A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $... | We claim that the smallest $n$ is $67^2 = \boxed{4489}$ | 489 |
1,645 | https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1 | 2 | A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $... | This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as "equivalence relation":
It is possible to write all positive integers $n$ in the form $p\cdot m^2$ , where $m^2$ is the largest perfect square dividing $n$ , so $p$ is not divisible by the square of any prime. Ob... | 489 |
1,646 | https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1 | 3 | A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $... | It's well known that there exists $f(n)$ and $g(n)$ such that $n = f(n) \cdot g(n)$ , no square divides $f(n)$ other than 1, and $g(n)$ is a perfect square.
We prove first: If $f(k) = f(a_k)$ $k \cdot a_k$ is a perfect square.
$k \cdot a_k = f(k) \cdot g(k) \cdot f(a_k) \cdot g(a_k) = f(k)^2 \cdot g(k) \cdot g(a_k)$ , ... | 489 |
1,647 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | 1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \[12x+18x=45\] Solving gives us $x=1.5$ . The $18x$ is Alicia so $18\times1.5=\boxed{27}$ | 27 |
1,648 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | 2 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | The relative speed of the two is $18+12=30$ , so $\frac{3}{2}$ hours would be required to travel $45$ miles. $d=st$ , so $x=18\cdot\frac{3}{2}=\boxed{27}$ | 27 |
1,649 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | 3 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | Since $18$ mph is $\frac{3}{2}$ times $12$ mph, Alicia will travel $\frac{3}{2}$ times as far as Beth. If $x$ is the distance Beth travels, \[\frac{3}{2}x+x=45\] \[\frac{5}{2}x=45\] \[x=18\] Since this is the amount Beth traveled, the amount that Alicia traveled was \[45-18=\boxed{27}\] | 27 |
1,650 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | 4 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | Alice and Barbara close in on each other at 30mph. Since they are 45 miles apart, they will meet in t = d/s = 45miles / 30mph = 3/2 hours. We can either calculate the distance Alice travels at 18mph or the distance Barbara travels at 12mph; since we want the distance from Alice, we go with the former. Alice (and Barbar... | 27 |
1,651 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | 5 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | We know that Beth starts 45 miles away from City A, let’s create two equations:
Alice-> $18t=d$ Beth-> $-12t+45=d$ [-12 is the slope; 45 is the y-intercept]
Solve the system:
$18t=-12t+45 30t=45 t=1.5$
So, $18(1.5)=$ $\boxed{27}$ | 27 |
1,652 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | 6 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | Since Alicia and Beth's speeds are constant, they are directly proportional to their distances covered, so the ratio of their speeds is equal to the ratio of their covered distances. Since Alicia travels $\frac{18}{30} = \frac{3}{5}$ of their combined speed, she travels $\frac{3}{5}\cdot 45 = \boxed{27}$ of the total d... | 27 |
1,653 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3 | 1 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | Note that $40^2=1600$ but $45^{2}=2025$ (which is over our limit of $2023$ ). Therefore, the list is $5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2$ . There are $8$ elements, so the answer is $\boxed{8}$ | 8 |
1,654 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3 | 4 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | We know the highest value would be at least $40$ but less than $50$ so we check $45$ , prime factorizing 45. We get $3^2 \cdot 5$ . We square this and get $81 \cdot 25$ . We know that $80 \cdot 25 = 2000$ , then we add 25 and get $2025$ , which does not satisfy our requirement of having the square less than $2023$ . Th... | 8 |
1,655 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | 1 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | Let's use the triangle inequality. We know that for a triangle, the sum of the 2 shorter sides must always be longer than the longest side. This is because if the longest side were to be as long as the sum of the other sides, or longer, we would only have a line.
Similarly, for a convex quadrilateral, the sum of the sh... | 12 |
1,656 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | 2 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | Say the chosen side is $a$ and the other sides are $b,c,d$
By the Generalised Polygon Inequality, $a<b+c+d$ . We also have $a+b+c+d=26\Rightarrow b+c+d=26-a$
Combining these two, we get $a<26-a\Rightarrow a<13$
The largest length that satisfies this is $a=\boxed{12}$ | 12 |
1,657 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | 3 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | The quadrilateral can by cyclic only when it is an isosceles triangle. Without Loss of Generality, lets assume that this quadrilateral is a trapezoid. We can assume this as if we inscribe a trapezoid in a triangle, the base can be the diameter of the circle which is the longest chord in the circle, therefore maximizing... | 12 |
1,658 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | 4 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | This is an AMC 10 problem 4, so there is no need for any complex formulas. The largest singular side length from a quadrilateral comes from a trapezoid. So we can set the $2$ sides of the trapezoid equal to $4$ . Next we can split the trapezoid into $5$ triangles, where each base length of the triangle equals $4$ . So ... | 12 |
1,659 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_5 | 1 | How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$
$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$ | Prime factorizing this gives us $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$
$10^{15}$ gives us $15$ digits and $243$ gives us $3$ digits. $15+3=\text{\boxed{18}$ | 18 |
1,660 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | 1 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | Each of the vertices is counted $3$ times because each vertex is shared by three different edges.
Each of the edges is counted $2$ times because each edge is shared by two different faces.
Since the sum of the integers assigned to all vertices is $21$ , the final answer is $21\times3\times2=\boxed{126}$ | 126 |
1,661 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | 2 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | Note that each vertex is counted $2\times 3=6$ times. Thus, the answer is $21\times6=\boxed{126}$ | 126 |
1,662 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | 3 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | Just set one vertice equal to $21$ , it is trivial to see that there are $3$ faces with value $42$ , and $42 \cdot 3=\boxed{126}$ | 126 |
1,663 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | 4 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | Since there are 8 vertices in a cube, there are $\dfrac{21}4$ vertices for two edges. There are $4$ edges per face, and $6$ faces in a cube, so the value of the cube is $\dfrac{21}4 \cdot 24 = \boxed{126}$ | 126 |
1,664 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | 5 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | Set each vertex to value 1, so the sum of the vertices is 8. We find that the value of the cube, if all vertices are 1, is 48. We conclude that the value of the cube is 6 times the value of the sum of the vertices. Therefore, we choose $21\times6=\boxed{126}$ | 126 |
1,665 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | 6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | The wording of the problem implies that the answer should hold for any valid combination of integers. Thus, we choose the numbers $21, 0, 0, 0, 0, 0, 0, 0$ , which are indeed $8$ integers that add to $21$ . Doing this, we find three edges that have a value of $21$ , and from there, we get three faces with a value of $4... | 126 |
1,666 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_9 | 2 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the... | There is one $3$ , so we need one more (three more means that either the month or units digit of the day is $3$ ). For the same reason, we need one more $0$
If $3$ is the units digit of the month, then the $0$ can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two... | 9 |
1,667 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_9 | 3 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the... | We start with $2023----$ we need an extra $0$ and an extra $3$ . So we have at least one of those extras in the days, except we can have the month $03$ . We now have $6$ possible months $01,02,03,10,11,12$ . For month $1$ we have two cases, we now have to add in another 1, and the possible days are $13,31$ . For month ... | 9 |
1,668 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | 1 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\q... | Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously.
We can write the following equations:
\[\frac{ax+11}{a+1}=x+1\qquad (1)\] \[\frac{ax+33}{a+3}=x+2\qquad (2)\]
Multiplying $(x+1)$ by $(a+1)$ and solving, we get: \[ax+11=ax+a+x+1\] \[11=a+x+1\] \[a+x=10\qquad (3)\]
M... | 7 |
1,669 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | 2 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\q... | Suppose Maureen took $n$ tests with an average of $m$
If she takes another test, her new average is $\frac{(nm+11)}{(n+1)}=m+1$
Cross-multiplying: $nm+11=nm+n+m+1$ , so $n+m=10$
If she takes $3$ more tests, her new average is $\frac{(nm+33)}{(n+3)}=m+2$
Cross-multiplying: $nm+33=nm+2n+3m+6$ , so $2n+3m=27$
But $2n+3m$ ... | 7 |
1,670 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | 3 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\q... | Let $s$ represent the sum of Maureen's test scores previously and $t$ be the number of scores taken previously.
So, $\frac{s+11}{t+1} = \frac{s}{t}+1$ and $\frac{s+33}{t+3} = \frac{s}{t}+2$
We can use the first equation to write $s$ in terms of $t$
We then substitute this into the second equation: $\frac{-t^2+10t+33}{t... | 7 |
1,671 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_12 | 1 | How many three-digit positive integers $N$ satisfy the following properties?
$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$ | Multiples of $5$ will always end in $0$ or $5$ , and since the numbers have to be a three-digit numbers (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with $5$ . Since the numbers must be divisible by 7, all possibilities have to be in the range fro... | 14 |
1,672 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_12 | 4 | How many three-digit positive integers $N$ satisfy the following properties?
$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$ | The key point is that when reversed, the number must start with a $0$ or a $5$ based on the second restriction. But numbers can't start with a $0$
So the problem is simply counting the number of multiples of $7$ in the $500$ s.
$7 \times 72 = 504$ , so the first multiple is $7 \times 72$
$7 \times 85 = 595$ , so the la... | 14 |
1,673 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_15 | 1 | An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed bel... | Notice that the area of the shaded region is $(2^2\pi-1^2\pi)+(4^2\pi-3^2\pi)+(6^2\pi-5^2\pi)+ \cdots + (n^2\pi-(n-1)^2 \pi)$ for any even number $n$
Using the difference of squares, this simplifies to $(1+2+3+4+\cdots+n) \pi$ . So, we are basically finding the smallest $n$ such that $\frac{n(n+1)}{2}>2023 \Rightarrow ... | 64 |
1,674 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_15 | 2 | An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed bel... | After first observing the problem, we can work out a few of the areas.
1st area = $4\pi-\pi = 3\pi$
2nd area = $16\pi-9\pi = 7\pi$
3rd area = $36\pi-25\pi = 11\pi$
4th area = $64\pi-49\pi = 15\pi$
We can see that the pattern is an arithmetic sequence with first term $3$ and common difference $4$ . From here, we can sta... | 64 |
1,675 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_15 | 3 | An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed bel... | We can easily see that all of the answer choices are even, which helps us solve this problem a little.
Lets just not consider the $\pi$ , since it is not that important, so let's just cancel that out.
When we plug in 64, we get $64^2-63^2+62^2-61^2+\cdots +4^2-3^2+2^2-1^2$ . By difference of squares, we get $1+2+3+\cdo... | 64 |
1,676 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_15 | 4 | An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed bel... | We can consider making a table.
If there is 1 circle, the area of the shaded region is 0π. (We can write this as 0π.)
If there are 2 circles, the area of the shaded region is 3π. (We can write this as (1+2)π).
If there are 3 circles, the area of the shaded region is 3π. (We can write this as (1+2)π).
If there are 4 c... | 64 |
1,677 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_15 | 5 | An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed bel... | Denote $S$ the area of shaded region and $W$ the area of white region.
$S > W$ and $S \approx W$ if $R$ is big.
Therefore \[\pi R^2 = S + W \approx 2S \ge 4046 \pi \implies R \approx 64.\] So we conclude the answer is $\boxed{64}$ | 64 |
1,678 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_16 | 1 | In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambid... | We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$ , and since $l = 1.4r$ $g = 2.4r$ . Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only answers ... | 36 |
1,679 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_16 | 2 | In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambid... | First, we know that every player played every other player, so there's a total of $\dbinom{n}{2}$ games since each pair of players forms a bijection to a game. Therefore, that rules out D. Also, if we assume the right-handed players won a total of $x$ games, the left-handed players must have won a total of $\dfrac{7}{5... | 36 |
1,680 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_16 | 3 | In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambid... | Let $r$ be the amount of games the right-handed won. Since the left-handed won $1.4r$ games, the total number of games played can be expressed as $(2.4)r$ , or $12/5r$ , meaning that the answer is divisible by 12. This brings us down to two answer choices, $B$ and $D$ .
We note that the answer is some number $x$ choos... | 36 |
1,681 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_16 | 4 | In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambid... | If there are $n$ players, the total number of games played must be $\binom{n}{2}$ , so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is $7:5$ , so the number of games played must also be divisible by $12$ . Finally, we notice that only $\boxed{36}$ satisfies both of the... | 36 |
1,682 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_18 | 2 | A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | Let $x$ be the number of vertices with three edges, and $y$ be the number of vertices with four edges. Since there are $\frac{4*12}{2}=24$ edges on the polyhedron, we can see that $\frac{3x+4y}{2}=24$ . Then, $3x+4y=48$ . Notice that by testing the answer choices, (D) is the only one that yields an integer solution for... | 8 |
1,683 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_18 | 3 | A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | With $12$ rhombi, there are $4\cdot12=48$ total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have $\dfrac{48}2=24$ total edges.
Let $A$ be the number of vertices with $3$ edges (this is what the problem asks for) and $B$ be the number of vertices with $4$ edges. We have ... | 8 |
1,684 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_18 | 4 | A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | Note that Euler's formula is $V+F-E=2$ . We know $F=12$ from the question. We also know $E = \frac{12 \cdot 4}{2} = 24$ because every face has $4$ edges and every edge is shared by $2$ faces. We can solve for the vertices based on this information.
Using the formula we can find: \[V + 12 - 24 = 2\] \[V = 14\] Let $t$ ... | 8 |
1,685 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_18 | 6 | A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | Note that a rhombic dodecahedron is formed when a cube is turned inside out (as seen here ), thus there are 6 4-vertices (corresponding to each face of the cube) and 8 3-vertices (corresponding to each corner of the cube). Thus the answer is $\boxed{8}$ | 8 |
1,686 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_18 | 7 | A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | Let $m$ be the number of $4$ -edge vertices, and $n$ be the number of $3$ -edge vertices. The total number of vertices is $m+n$ . Now, we know that there are $4 \cdot 12 = 48$ vertices, but we have overcounted. We have overcounted $m$ vertices $3$ times and overcounted $n$ vertices $2$ times. Therefore, we subtract $3m... | 8 |
1,687 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_19 | 1 | The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$ | Due to rotations preserving an equal distance, we can bash the answer with the distance formula. $D(A, P) = D(A', P)$ , and $D(B, P) = D(B',P)$ .
Thus we will square our equations to yield: $(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2$ , and $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$ .
Canceling $(3-s)^2$ from the second equation makes it ... | 1 |
1,688 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_19 | 2 | The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$ | Due to rotations preserving distance, we have that $BP = B^\prime P$ , as well as $AP = A^\prime P$ . From here, we can see that P must be on the perpendicular bisector of $\overline{BB^\prime}$ due to the property of perpendicular bisectors keeping the distance to two points constant.
From here, we proceed to find the... | 1 |
1,689 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_19 | 3 | The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$ | To find the center of rotation, we find the intersection point of the perpendicular bisectors of $\overline{AA^\prime}$ and $\overline{BB^\prime}$
We can find that the equation of the line $\overline{AA^\prime}$ is $y = -\frac{1}{2}x + \frac{5}{2}$ , and that the equation of the line $\overline{BB^\prime}$ is $y = 3$
W... | 1 |
1,690 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_19 | 4 | The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$ | We use the complex numbers approach to solve this problem.
Denote by $\theta$ the angle that $AB$ rotates about $P$ in the counterclockwise direction.
Thus, $A' - P = e^{i \theta} \left( A - P \right)$ and $B' - P = e^{i \theta} \left( B - P \right)$
Taking ratio of these two equations, we get \[ \frac{A' - P}{A - P} =... | 1 |
1,691 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20 | 1 | Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)... | Let a "tile" denote a $1\times1$ square and "square" refer to $2\times2$
We first have $4!=24$ possible ways to fill out the top left square. We then fill out the bottom right tile. In the bottom right square, we already have one corner filled out (from our initial coloring), and we now have $3$ options left to pick fr... | 72 |
1,692 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20 | 2 | Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)... | [asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("R", (1.5,1.5)); label("B", (4.5,1.5)); label("R", (7.5,1.5)); label("G", (1.5,4.5)); label("W", (4.5,4.5)); label("G", (7.5,4.5)); label("B", (1.5,7.5)); label... | 72 |
1,693 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20 | 3 | Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)... | Let’s call the top-right corner color A, the top-middle color B, the top-right color C, and so on, with color D being the middle row, and right corner square, and color G being the bottom-left square’s color. WLOG A=Red, B=White, D=Blue, and E=Green. We will now consider squares C and F’s colors.
Case 1 : C=Red and F=B... | 72 |
1,694 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20 | 4 | Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)... | We will choose colors step-by-step:
1. There are $4$ ways to choose a color in the center.
2. Then we select any corner and there would be $3$ ways to choose a color as we can't use the same color as the one in the center.
3. Consider the $2\times 2$ square that contains the center and the corner we have selected. For ... | 72 |
1,695 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20 | 5 | Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)... | Note that there can be no overlap between colors in each square.
Then, we can choose $1$ color to be in the center. ${4 \choose 1}$ = 4
Now, we have some casework:
Case 1: 1 color is placed in 4 corners and then others are placed on opposite edges. $232$ $414$ $232$ There's $3!=6$ ways to do this.
Case 2: 2 colors are ... | 72 |
1,696 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20 | 6 | Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)... | [asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("2", (1.5,1.5)); label("1", (4.5,1.5)); label("1", (7.5,1.5)); label("2", (1.5,4.5)); label("3", (4.5,4.5)); label("1", (7.5,4.5)); label("3", (1.5,7.5)); label... | 72 |
1,697 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_21 | 1 | Let $P(x)$ be the unique polynomial of minimal degree with the following properties:
The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. What is $m+n$
$\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(... | From the problem statement, we know $P(2-2)=0$ $P(9)=0$ and $4P(4)=0$ . Therefore, we know that $0$ $9$ , and $4$ are roots. So, we can factor $P(x)$ as $x(x - 9)(x - 4)(x - a)$ , where $a$ is the unknown root. Since $P(x) - 1 = 0$ , we plug in $x = 1$ which gives $1(-8)(-3)(1 - a) = 1$ , therefore $24(1 - a) = 1 \impl... | 47 |
1,698 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_21 | 2 | Let $P(x)$ be the unique polynomial of minimal degree with the following properties:
The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. What is $m+n$
$\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(... | We proceed similarly to solution one. We get that $x(x-9)(x-4)(x-a)=1$ . Expanding, we get that $x(x-9)(x-4)(x-a)=x^4-(a+13)x^3+(13a+36)x^2-36ax$ . We know that $P(1)=1$ , so the sum of the coefficients of the cubic expression is equal to one. Thus $1+(a+13)+(13a+36)-36a=1$ . Solving for a, we get that a=23/24. Therefo... | 47 |
1,699 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_23 | 2 | If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$ , then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$ . Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ b... | We have 4 integers in our problem. Let's call the smallest of them $a$ $a(a+23) =$ either $(a+1)(a+21)$ or $(a+2)(a+22)$ . So, we have the following:
$a^2 + 23a = a^2 + 22a +21$ or
$a^2+23a = a^2 + 24a +44$
The second equation has negative solutions, so we discard it. The first equation has $a = 21$ , and so $a + 23 = ... | 15 |
1,700 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_23 | 3 | If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$ , then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$ . Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ b... | From the problems, it follows that
\begin{align*} x(x+20)&=y(y+23) = N\\ x^2+20x&=y^2+23y\\ 4x^2+4\cdot20x &= 4y^2+4\cdot23y\\ 4x^2+4\cdot20x+20^2-20^2 &= 4y^2+4\cdot23y+23^2-23^2\\ (2x+20)^2-20^2 &= (2y+23)^2-23^2\\ 23^2-20^2 &= (2y+23)^2-(2x+20)^2\\ (23+20)(23-20) &= (2y+23+2x+20)(2y+23-2x-20)\\ 43\cdot 3 &= (2y+2x+4... | 15 |
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