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 |
|---|---|---|---|---|---|
5,901 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9 | 2 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=or... | Visualizing the corner which is submerged in the cylinder, we can see that it's like slicing off a corner, where they slice an equal part off every edge, to make a tetrahedron where there are 3 right isosceles triangles and one equilateral triangle. With this out of the way, we can now just find the area of that equila... | 384 |
5,902 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9 | 3 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=or... | We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\sqrt3$ . From here, its area is \[\dfrac{\bigl(4\sqrt3\bigr)^2\sqrt3}4=12\sqrt3.\] The leg of the isosceles right triangle is $\dfrac{4\sqrt3}{\sqrt2}=2\sqrt6$ , and the horizontal distance from the vertex to th... | 384 |
5,903 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9 | 4 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=or... | Since the diagonal is perpendicular to the base of the cylinder, all three edges and faces can be treated symmetrically.
The cross-section of the cube with the top face of the cylinder is an equilateral triangle inscribed in a circle with radius $4$ . This means the medians of the triangle have length $\frac{3}{2} \cdo... | 384 |
5,904 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_10 | 1 | Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$ . For example, 53421 and 14253 are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$ , but 45123 is not. Find the number of quasi-increasing permutations of th... | The simple recurrence can be found.
When inserting an integer $n$ into a string with $n - 1$ integers, we notice that the integer $n$ has 3 spots where it can go: before $n - 1$ , before $n - 2$ , and at the very end.
EXAMPLE:
Putting 4 into the string 123:
4 can go before the 2: 1423,
Before the 3: 1243,
And at the v... | 486 |
5,905 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_11 | 5 | The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | $\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}$
Denote the circumradius of $ABC$ to be $R$ , the circumcircle of $ABC$ to be $O$ , and the shortest distance from $Q$ to circle $O$ to be $x$
Using Power of a Point on $Q$ relative to circle $O$ , we get th... | 23 |
5,906 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | 1 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | Let $a_{n}$ be the number of ways to form $n$ -letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical.
Note that, at the end of each $n$ -letter string, there are $3$ possibilities for the last letter chain: it must be either $1$ $2$ , or $3$ letters long. Removing this last chain... | 548 |
5,907 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | 2 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | The solution is a simple recursion:
We have three cases for the ending of a string: three in a row, two in a row, and a single:
...AAA $(1)$ ...BAA $(2)$ ...BBA or ...ABA $(3)$
(Here, WLOG each string ends with A. This won't be the case when we actually solve for values in recursion.)
For case $(1)$ , we could only add... | 548 |
5,908 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | 3 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | This is a recursion problem. Let $a_n$ be the number of valid strings of $n$ letters, where the first letter is $A$ . Similarly, let $b_n$ be the number of valid strings of $n$ letters, where the first letter is $B$
Note that $a_n=b_{n-1}+b_{n-2}+b_{n-3}$ for all $n\ge4$
Similarly, we have $b_n=a_{n-1}+a_{n-2}+a_{n-3}... | 548 |
5,909 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | 4 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | Playing around with strings gives this approach: We have a certain number of As, then Bs, and so on. Therefore, what if we denoted each solution with numbers like $(3,3,3,1)$ to denote AAABBBAAAB or vice versa (starting with Bs)? Every string can be represented like this!
We can have from 4 to 10 numbers in our parenth... | 548 |
5,910 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | 5 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | We are going to build the string, 1 character at a time. And, we are going to only care about the streak of letters at the end of the string.
Let $a_n$ be the number of strings of length n that satisfy the problem statement and also has a "streak" of length 1 at the end. ABABBA has a streak of length 1.
Let $b_n$ be th... | 548 |
5,911 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | 6 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | Let $S_n$ be the number of n-letter strings satisfying the problem criteria. Then we can easily see $S_1 = 2, S_2 = 4, S_3 = 8, S_4 = 16-2 =14$ . For $n \ge 4$ consider the last three elements of the list. For example to get $S_5$ we could try to take $S_4$ and multiply by 2 since for each $S_4$ string we have two choi... | 548 |
5,912 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_13 | 1 | Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$ | If $n = 1$ $a_n = \sin(1) > 0$ . Then if $n$ satisfies $a_n < 0$ $n \ge 2$ , and \[a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{2\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{2\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].\] Since $2\sin 1$ is positive, it ... | 628 |
5,913 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_13 | 2 | Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$ | Notice that $a_n$ is the imaginary part of $\sum_{k=1}^n e^{ik}$ , by Euler's formula. Using the geometric series formula, we find that this sum is equal to \[\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}\] We multiply the fraction by the conjugate of the de... | 628 |
5,914 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_13 | 3 | Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$ | Similar to solution 2, we set a complex number $z=\cos 1+i\sin 1$ . We start from $z$ instead of $1$ because $k$ starts from $1$ : be careful.
The sum of $z+z^2+z^3+z^4+z^5\dots=\frac{z-z^{n+1}}{1-z}=\frac{z}{z-1}\left(z^n-1\right)$
We are trying to make $n$ so that the imaginary part of this expression is negative.
Th... | 628 |
5,915 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | 7 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | Take $w=x+y$ and $z=xy$ . Remark that \begin{align*} ~&2x^{3}+(xy)^{3}+2y^{3} \\ =~&2(x^{3}+y^{3})+(xy)^{3} \\ =~&2\left[(x+y)^{3}-3xy(x+y)\right]+(xy)^{3} \\ =~&2(w^{3}-3wz)+z^{3} \\ =~& 2w^{3}-6wz+z^{3}.\end{align*} The given equations imply that \[wz^{4}=810~~~\text{and}~~~(wz)^{3}-3wz^{4}=945.\] Substituting the fi... | 89 |
5,916 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | 8 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | $x^4y^4(x+y)=810; x^3y^3(x^3+y^3)=945, \frac{x^2-xy+y^2}{xy}=\frac{7}{6}, \frac{x^2+y^2}{xy}=\frac{13}{6}$
Let $x^2+y^2=13k; xy=6k$ , then we can see $(x+y)^2-12k=13k, x+y=5\sqrt{k}$ , now, we see $x^4y^4\cdot (x+y)=1296k^4\cdot 5\sqrt{k}, k=\frac{1}{\sqrt[3]{4}}$
The rest is easy, $2(x^3+y^3)+x^3y^3=216k^3+2[(x+y)^3-3... | 89 |
5,917 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | 9 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | Let's first put the left hand sides of the equations in factored forms. Doing this we obtain $(xy)^4 \cdot (x + y) = 810$ and $(xy)(x + y)(x^2 - xy + y^2) = 945$ . Now, we will subtract and add the equations to gather information on $x$ and $y$ . When we subtract the equations and clean it up via factoring, we yield $(... | 89 |
5,918 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | 1 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | Let $M$ be the intersection of $\overline{BC}$ and the common internal tangent of $\mathcal P$ and $\mathcal Q.$ We claim that $M$ is the circumcenter of right $\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\overline{BC},$ implying that... | 129 |
5,919 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | 2 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | [asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label("$A$",A,NE);label("$O_1$",O_... | 129 |
5,920 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | 3 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | By homothety , we deduce that $AE = 4 AD$ . (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of $P$ and $Q$ to $l$ .) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from $B$ to $l$ ... | 129 |
5,921 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | 4 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | $DE$ goes through $A$ , the point of tangency of both circles. So $DE$ intercepts equal arcs in circle $P$ and $Q$ homothety . Hence, $AE=4AD$ . We will use such similarity later.
The diagonal distance between the centers of the circles is $4+1=5$ . The difference in heights is $4-1=3$ . So $BC=\sqrt{5^2-3^2}=4$
The tr... | 129 |
5,922 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | 5 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | Let $K$ be the intersection of $BC$ and $AE$ . Since the radii of the two circles are 1:4, so we have $AD:AE=1:4$ , and the distance from $B$ to line $l$ and the distance from $C$ to line $l$ are in a ratio of 4:1, so $BK:CK=4:1$ . We can easily calculate the length of $BC$ to be 4, so $CK=\frac{4}{3}$ . Let $J$ be the... | 129 |
5,923 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | 6 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | Consider the common tangent from $A$ to both circles. Let this intersect $BC$ at point $K$ . From equal tangents, we have $BK=AK=CK$ , which implies that $\angle BAC = 90^\circ$
Let the center of $\mathcal{P}$ be $O_1$ , and the center of $\mathcal{Q}$ be $O_2$ . Angle chasing, we find that $\triangle O_1DA \sim \trian... | 129 |
5,924 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | 7 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | Add in the line $k$ as the internal tangent between the two circles. Let $M$ be the midpoint of $BC$ ; It is well-known that $M$ is on $k$ and because $k$ is the radical axis of the two circles, $AM=BM=CM$ . Therefore because $M$ is the circumcenter of $\triangle{BAC}$ $\angle{BAC}=90^{\circ}$ . Let $O_P$ be the center... | 129 |
5,925 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | 8 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | Let $P$ and $Q$ be the centers of circles $\mathcal{P}$ and $\mathcal{Q}$ , respectively.
Let $M$ be midpoint $BC, \beta = \angle ACB.$
Upper diagram shows that
$\sin 2\beta = \frac {4}{5}$ and $AC = 2 AB.$ Therefore $\cos 2\beta = \frac {3}{5}.$
Let $CH\perp l, BH'\perp l.$ Lower diagram shows that
$\angle CAE = \angl... | 129 |
5,926 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | 9 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | We begin by extending $\overline{AB}$ upwards until it intersects Circle $\mathcal{Q}$ . We can call this point of intersection $F$ . Connect $F$ with $E$ $C$ , and $A$ for future use.
Create a trapezoid with points $B$ $C$ , and the origins of Circles $\mathcal{P}$ and $\mathcal{Q}$ . After quick inspection, we can co... | 129 |
5,927 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | 10 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | Let the center of the larger circle be $O,$ and the center of the smaller circle be $P.$ It is not hard to find the areas of $ACO$ and $ABP$ using pythagorean theorem, which are $\frac{32}{5}$ and $\frac{2}{5}$ respectively. Assign $\angle AOC=a,\angle COE=c,\angle DPB=d,\angle BPA=b.$ We can figure out that $\angle AO... | 129 |
5,928 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_1 | 1 | The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and the... | The rectangle is divided into three smaller rectangles with a width of 50 mm and a length of $\dfrac{80}{3}$ mm. According to the Pythagorean Theorem (or by noticing the 8-15-17 Pythagorean triple), the diagonal of the rectangle is $\sqrt{50^2+\left(\frac{80}{3}\right)^2}=\frac{170}{3}$ mm. Since that on the lace, th... | 790 |
5,929 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_2 | 1 | An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$ . Find $N$ | First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to $0.58$
The probability both are green is $\frac{4}{10}\cdot\frac{16}{16+N}$ , and the probability both are blue is $\frac{6}{10}\cdot\frac{N}{16+N}$ , so \[\frac{4}{10}\cdot\frac{... | 144 |
5,930 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | 1 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | Let the numerator and denominator $x,y$ with $\gcd(x,y)=1$ and $x+y = 1000.$ Now if $\gcd(x,y) = 1$ then $\gcd(x,y) =\gcd(x,1000-x)= \gcd(x,1000-x-(-1)x)=\gcd(x,1000)=1.$ Therefore any pair that works satisfies $\gcd(x,1000)= 1.$ By Euler's totient theorem, there are $\phi(1000) = 400$ numbers relatively prime to 1000 ... | 200 |
5,931 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | 2 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | If the initial manipulation is not obvious, consider the Euclidean Algorithm. Instead of using $\frac{n}{m}$ as the fraction to use the Euclidean Algorithm on, we can rewrite this as $\frac{500-x}{500+x}$ or \[\gcd(500+x,500-x)=\gcd((500+x)+(500-x),500-x)=\gcd(1000,500-x).\] Thus, we want $\gcd(1000,500-x)=1$ . You can... | 200 |
5,932 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | 3 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | Say $r=\frac{d}{1000-d}$ ; then $1\leq d\leq499$ . If this fraction is reducible, then the modulus of some number for $d$ is the same as the modulus for $1000-d$ . Since $1000=2^3\cdot5^3$ , that modulus can only be $2$ or $5$ . This implies that if $d\mid2$ or $d\mid5$ , the fraction is reducible. There are $249$ case... | 200 |
5,933 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | 4 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | We know that the numerator of the fraction cannot be even, because the denominator would also be even. We also know that the numerator cannot be a multiple of $5$ because the denominator would also be a multiple of $5$ . Proceed by listing out all the other possible fractions and we realize that the numerator and denom... | 200 |
5,934 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | 5 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | We know that the set of these rational numbers is from $\dfrac{1}{999}$ to $\dfrac{499}{501}$ where each each element $\dfrac{n}{m}$ has $n+m =1000$ and $\dfrac{n}{m}$ is irreducible.
We note that $\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1$ .
Hence, $\dfrac{n}{m}$ is irreducible if $\dfrac{1000}{m}$ is irreduci... | 200 |
5,935 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | 6 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | We notice that there are a total of $400$ fractions that are in simplest form where the numerator and denominator add up to $1000$ . Because the numerator and denominator have to be relatively prime, there are $\varphi(1000)=400$ fractions. Half of these are greater than $1$ , so the answer is $400\div2=\boxed{200}$ | 200 |
5,936 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | 7 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | Our fraction can be written in the form $\frac{1000 - a}{a} = \frac{1000}{a} - 1.$ Thus the fraction is reducible when $a$ divides $1000.$ We also want $500 < a < 1000.$ By PIE , the total values of $a$ that make the fraction reducible is, \[249 + 99 - 49 = 299.\] By complementary counting , the answer we want is $499 ... | 200 |
5,937 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | 8 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | Suppose our fraction is $\frac{a}{b}$ . The given condition means $a+b=1000$ . Now, if $a$ and $b$ share a common factor greater than $1$ , then the expression $a+b$ must also contain that common factor. This means our fraction cannot have a factor of $5$ or be even.
There are $250$ fractions that aren’t even. From thi... | 200 |
5,938 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_5 | 1 | Let the set $S = \{P_1, P_2, \dots, P_{12}\}$ consist of the twelve vertices of a regular $12$ -gon. A subset $Q$ of $S$ is called "communal" if there is a circle such that all points of $Q$ are inside the circle, and all points of $S$ not in $Q$ are outside of the circle. How many communal subsets are there? (Note tha... | By looking at the problem and drawing a few pictures, it quickly becomes obvious that one cannot draw a circle that covers $2$ disjoint areas of the $12$ -gon without including all the vertices in between those areas. In other words, in order for a subset to be communal, all the vertices in the subset must be adjacent ... | 134 |
5,939 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6 | 5 | The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ | We have the equation $y=3(x-h)^2 + j.$
We know: $(x,y):(0,2013)$ , so $h^2=2013/3 - j/3$ after plugging in the values and isolating $h^2$ . Therefore, $h^2=671-j/3$
Lets call the x-intercepts $x_1$ $x_2$ . Since both $x_1$ and $x_2$ are positive there is a relationship between $x_1$ $x_2$ and $h$ . Namely, $x_1+x_2=2h$... | 36 |
5,940 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6 | 6 | The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ | First, we start of exactly like solutions above and we find out that $j=2013-3h^2$ and $k=2014-2h^2$ We then plug j and k into $3(x-h)^2+j$ and $y=2(x-h)^2+k$ respectively. After that, we get two equations, $y=3x^2-6xh+2013$ and $y=2x^2-4xh+2014$ . We can apply Vieta's. Let the roots of the first equation be $a, b$ and... | 36 |
5,941 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_7 | 1 | Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denote... | Let $w = \operatorname{cis}{(\alpha)}$ and $z = 10\operatorname{cis}{(\beta)}$ . Then, $\dfrac{w - z}{z} = \dfrac{\operatorname{cis}{(\alpha)} - 10\operatorname{cis}{(\beta)}}{10\operatorname{cis}{\beta}}$
Multiplying both the numerator and denominator of this fraction by $\operatorname{cis}{(-\beta)}$ gives us:
$\dfra... | 100 |
5,942 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_7 | 2 | Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denote... | Without the loss of generality one can let $z$ lie on the positive x axis and since $arg(\theta)$ is a measure of the angle if $z=10$ then $arg(\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivalent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$... | 100 |
5,943 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_7 | 3 | Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denote... | Note that $\frac{w-z}{z}=\frac{w}{z}-1$ , and that $\left|\frac{w}{z}\right|=\frac{1}{10}$ . Thus $\frac{w}{z}-1$ is a complex number on the circle with radius $\frac{1}{10}$ and centered at $-1$ on the complex plane. Let $\omega$ denote this circle.
Let $A$ and $C$ be the points that represent $\frac{w}{z}-1$ and $-1$... | 100 |
5,944 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8 | 1 | The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$ | We have that $N^2 - N = N(N - 1)\equiv 0\mod{10000}$
Thus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$ . Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$ , which is impossible for a number that is divisible by either $2$ or $5$ . Th... | 937 |
5,945 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8 | 2 | The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$ | let $N= 10000t+1000a+100b+10c+d$ for positive integer values $t,a,b,c,d$ .
When we square $N$ we get that \begin{align*} N^2 &=(10000t+1000a+100b+10c+d)^2\\ &=10^8t^2+10^6a^2+10^4b^2+10^2c^2+d^2+2(10^7ta+10^6tb+10^5tc+10^4td+10^5ab+10^4ac+10^3bc+10^ad+10^2bd+10cd) \end{align*}
However, we don't have to deal with this w... | 937 |
5,946 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8 | 3 | The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$ | By the Chinese Remainder Theorem, the equation $N(N-1)\equiv 0\pmod{10000}$ is equivalent to the two equations: \begin{align*} N(N-1)&\equiv 0\pmod{16},\\ N(N-1)&\equiv 0\pmod{625}. \end{align*} Since $N$ and $N-1$ are coprime, the only solutions are when $(N\mod{16},N\mod{625})\in\{(0,0),(0,1),(1,0),(1,1)\}$
Let \[\va... | 937 |
5,947 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8 | 4 | The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$ | WLOG, we can assume that $N$ is a 4-digit integer $\overline{abcd}$ . Note that the only $d$ that will satisfy $N$ will also satisfy $d^2\equiv d\pmod{10}$ , as the units digit of $\overline{abcd}^2$ is affected only by $d$ , regardless of $a$ $b$ , or $c$
By checking the numbers 0-9, we see that the only possible valu... | 937 |
5,948 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11 | 1 | A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a po... | Perform the coordinate transformation $(x, y)\rightarrow (x+y, x-y)$ . Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\langle 1, -1 \rangle$ $\langle 1, 1 \rangle$ $\langle -1, -1 \rangle$ $\langle -1, 1 \rangle$ respectively. Moreover, the transformation takes the equ... | 391 |
5,949 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11 | 2 | A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a po... | We have 4 possible moves: U, D, R, and L. The total number of paths that could be taken is $4^6$ , or $4096$ . There are 4 possible cases that land along the line $y = x$ $x,y = \pm 1; x,y = \pm 2; x,y = \pm 3;$ or $x = y = 0$ . We will count the number of ways to end up at $(1,1), (2,2),$ and $(3,3)$ , multiply them b... | 391 |
5,950 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11 | 3 | A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a po... | We split this into cases by making a chart. In each row, the entries $(\pm1)$ before the dividing line represent the right or left steps (without regard to order), and the entries after the dividing line represent the up or down steps (again, without regard to order). This table only represents the cases where the endi... | 391 |
5,951 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11 | 4 | A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a po... | Denote $(x, y)_n$ the probability that starting from point $(x, y)$ , the token reaches a point on the graph of $|y| = |x|$ in exactly $n$ moves. The problem asks us to find $(0, 0)_6$ . We start by breaking this down: \[(0, 0)_6 = \frac14 \cdot ((0, 1)_5 + (0, -1)_5 + (1, 0)_5 + (-1, 0)_5)\] Notice that by symmetry, $... | 391 |
5,952 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_12 | 1 | Let $A=\{1,2,3,4\}$ , and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$ . The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | The natural way to go is casework. And the natural process is to sort $f$ and $g$ based on range size! Via Pigeonhole Principle, we see that the only real possibilities are: $f 1 g 1; f 1 g 2; f 1 g 3; f 2 g 2$ . Note that the $1, 2$ and $1, 3$ cases are symmetrical and we need just a $*2$ . Note also that the total nu... | 453 |
5,953 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_12 | 2 | Let $A=\{1,2,3,4\}$ , and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$ . The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | We note there are $4^4 = 256$ possibilities for each of $f$ and $g$ from $A$ to $A$ since the input of the four values of each function has four options each for an output value.
We proceed with casework to determine the number of possible $f$ with range 1, 2, etc.
There are 4 possibilities: all elements output to 1, 2... | 453 |
5,954 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_13 | 1 | On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE,... | $269+275+405+411=1360$ , a multiple of $17$ . In addition, $EG=FH=34$ , which is $17\cdot 2$ .
Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to $EG$ and $FH$ must be a multiple of $17$ . All of these triples are primitive:
\[17=1^2+4^2\] \[34=3^2+5^2\] \[51=\emptyset\] \[68=\em... | 850 |
5,955 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_13 | 2 | On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE,... | Continue in the same way as solution 1 to get that $POK$ has area $3a$ , and $OK = \frac{d}{10}$ . You can then find $PK$ has length $\frac 32$
Then, if we drop a perpendicular from $H$ to $BC$ at $L$ , We get $\triangle HLF \sim \triangle OPK$
Thus, $LF = \frac{15\cdot 34}{d}$ , and we know $HL = d$ , and $HF = 34$ . ... | 850 |
5,956 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_14 | 1 | Let $m$ be the largest real solution to the equation
$\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4$
There are positive integers $a$ $b$ , and $c$ such that $m=a+\sqrt{b+\sqrt{c}}$ . Find $a+b+c$ | The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $\frac{3}{x-3}$ , then the fraction becomes of the form $\frac{x}{x - 3}$ . A similar cancellation happens with the other four terms. If we assume $x = 0$ is not the highest solution (which ... | 263 |
5,957 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_1 | 1 | Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together ... | From the given information, we can see that Abe can paint $\frac{1}{15}$ of the room in an hour, Bea can paint $\frac{1}{15}\times\frac{3}{2} = \frac{1}{10}$ of the room in an hour, and Coe can paint the room in $\frac{1}{15}\times 2 = \frac{2}{15}$ of the room in an hour. After $90$ minutes, Abe has painted $\frac{1}... | 334 |
5,958 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_3 | 1 | A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with... | When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below.
[asy] pair R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 2... | 720 |
5,959 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_3 | 2 | A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with... | Alternatively, use basic geometry. First, scale everything down by dividing everything by 6. Let $a/6=p$ . Then, the dimensions of the central rectangle in the hexagon is p x 4, and the original rectangle is 6 x p. By Pythagorean theorem and splitting the end triangles of the hexagon into two right triangles, the altit... | 720 |
5,960 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4 | 1 | The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ | Notice repeating decimals can be written as the following:
$0.\overline{ab}=\frac{10a+b}{99}$
$0.\overline{abc}=\frac{100a+10b+c}{999}$
where a,b,c are the digits. Now we plug this back into the original fraction:
$\frac{10a+b}{99}+\frac{100a+10b+c}{999}=\frac{33}{37}$
Multiply both sides by $999*99.$ This helps simpli... | 447 |
5,961 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4 | 2 | The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ | Note that $\frac{33}{37}=\frac{891}{999} = 0.\overline{891}$ . Also note that the period of $0.abab\overline{ab}+0.abcabc\overline{abc}$ is at most $6$ . Therefore, we only need to worry about the sum $0.ababab+ 0.abcabc$ . Adding the two, we get \[\begin{array}{ccccccc}&a&b&a&b&a&b\\ +&a&b&c&a&b&c\\ \hline &8&9&1&8... | 447 |
5,962 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4 | 3 | The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ | Noting as above that $0.\overline{ab} = \frac{10a + b}{99}$ and $0.\overline{abc} = \frac{100a + 10b + c}{999}$ , let $u = 10a + b$ .
Then \[\frac{u}{99} + \frac{10u + c}{999} = \frac{33}{37}\]
\[\frac{u}{11} + \frac{10u + c}{111} = \frac{9\cdot 33}{37}\]
\[\frac{221u + 11c}{11\cdot 111} = \frac{9\cdot 33}{37}\]
\[221u... | 447 |
5,963 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4 | 4 | The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ | We note as above that $0.\overline{ab} = \frac{10a + b}{99}$ and $0.\overline{abc} = \frac{100a + 10b + c}{999},$ so
\[\frac{10a + b}{99} + \frac{100a + 10b + c}{999} = \frac{33}{37} = \frac{891}{999}.\]
As $\frac{10a + b}{99}$ has a factor of $11$ in the denominator while the other two fractions don't, we need that $1... | 447 |
5,964 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5 | 1 | Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ | Because the coefficient of $x^2$ in both $p(x)$ and $q(x)$ is 0, the remaining root of $p(x)$ is $-(r+s)$ , and the remaining root of $q(x)$ is $-(r+s+1)$ . The coefficients of $x$ in $p(x)$ and $q(x)$ are both equal to $a$ , and equating the two coefficients gives \[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2\] from which $s = ... | 420 |
5,965 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5 | 2 | Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ | Let $r$ $s$ , and $-r-s$ be the roots of $p(x)$ (per Vieta's). Then $r^3 + ar + b = 0$ and similarly for $s$ . Also, \[q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0\]
Set up a similar equation for $s$
\[q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.\]
Simplifying and adding the eq... | 420 |
5,966 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5 | 3 | Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ | The roots of $p(x)$ are $r$ $s$ , and $-r-s$ since they sum to $0$ by Vieta's Formula (co-efficient of $x^2$ term is $0$ ).
Similarly, the roots of $q(x)$ are $r + 4$ $s - 3$ , and $-r-s-1$ , as they too sum to $0$
Then:
$a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2$ and $-b = rs(-r-s)$ from $p(x)$ and
$a=(r+4)(s-3)+(r+4)(-r-s-... | 420 |
5,967 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5 | 4 | Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ | By Vieta's, we know that the sum of roots of $p(x)$ is $0$ . Therefore,
the roots of $p$ are $r, s, -r-s$ . By similar reasoning, the roots of $q(x)$ are $r + 4, s - 3, -r - s - 1$ . Thus, $p(x) = (x - r)(x - s)(x + r + s)$ and $q(x) = (x - r - 4)(x - s + 3)(x + r + s + 1)$
Since $p(x)$ and $q(x)$ have the same coeffic... | 420 |
5,968 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_6 | 1 | Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ . Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixe... | The probability that he rolls a six twice when using the fair die is $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$ . The probability that he rolls a six twice using the biased die is $\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}$ . Given that Charles rolled two sixes, we can see that it is $16$ times more l... | 167 |
5,969 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_6 | 2 | Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ . Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixe... | This is an incredibly simple problem if one is familiar with conditional probability (SO BE FAMILIAR WITH IT)!
The conditional probability that the third roll will be a six given that the first two rolls are sixes, is the conditional probability that Charles rolls three sixes given that his first two rolls are sixes. ... | 167 |
5,970 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8 | 1 | Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the ... | Using the diagram above, let the radius of $D$ be $3r$ , and the radius of $E$ be $r$ . Then, $EF=r$ , and $CE=2-r$ , so the Pythagorean theorem in $\triangle CEF$ gives $CF=\sqrt{4-4r}$ . Also, $CD=CA-AD=2-3r$ , so \[DF=DC+CF=2-3r+\sqrt{4-4r}.\] Noting that $DE=4r$ , we can now use the Pythagorean theorem in $\triangl... | 254 |
5,971 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8 | 2 | Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the ... | Consider a reflection of circle $E$ over diameter $\overline{AB}$ . By symmetry, we now have three circles that are pairwise externally tangent and all internally tangent to a large circle. The small circles have radii $r$ $r$ , and $3r$ , and the big circle has radius $2$
Descartes' Circle Theorem gives $\left(\frac{1... | 254 |
5,972 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8 | 3 | Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the ... | We use the notation of Solution 1 for triangle $\triangle DEC$ \[\sin \angle EDC = \frac {EF}{DE} = \frac {1}{4} \implies \cos \angle EDC = \frac {\sqrt{15}}{4}.\] We use Cosine Law for $\triangle DEC$ and get: \[(4r)^2 +(2 – 3r)^2 – 2 \cdot 4r \cdot (2 – 3r) \cdot \frac {\sqrt{15}}{4} = (2 – r)^2\] \[(24 + 6 \sqrt{1... | 254 |
5,973 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_9 | 1 | Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. | We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs. There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent. There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10 \cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have all $5$ a... | 581 |
5,974 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_9 | 2 | Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. | Starting with small cases, we see that four chairs give $4 + 1 = 5$ , five chairs give $5 + 5 + 1 = 11$ , and six chairs give $6 + 6 + 6 + 6 + 1 = 25.$ Thus, n chairs should give $n 2^{n-4} + 1$ , as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are $n 2^{n-3}$ ways to arran... | 581 |
5,975 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_9 | 3 | Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. | It is possible to use recursion to count the complement. Number the chairs $1, 2, 3, ..., 10.$ If chair $1$ is not occupied, then we have a line of $9$ chairs such that there is no consecutive group of three. If chair $1$ is occupied, then we split into more cases. If chairs $2$ and $10$ are empty, then we have a line ... | 581 |
5,976 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_12 | 1 | Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$ | WLOG, let C be the largest angle in the triangle.
As above, we can see that $\cos3A+\cos3B-\cos(3A+3B)=1$
Expanding, we get
$\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1$
$\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B$
$(\cos3A-1)(\cos3B-1)=\sin3A\sin3B$
CASE 1: If $\sin 3A = 0$ or $\sin 3B = 0$
This implies one or both of A ... | 399 |
5,977 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_12 | 2 | Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$ | \[\cos3A+\cos3B=1-\cos(3C)=1+\cos(3A+3B)\] \[2\cos\frac{3}{2}(A+B)\cos\frac{3}{2}(A-B)=2\cos^2\frac{3}{2}(A+B)\] If $\cos\frac{3}{2}(A+B) = 0$ , then $\frac{3}{2}(A+B)=90$ $A+B=60$ , so $C=120$ ; otherwise, \[2\cos\frac{3}{2}(A-B)=2cos\frac{3}{2}(A+B)\] \[\sin\frac{3}{2}A\sin\frac{3}{2}B=0\] so either $\sin\frac{3}{2}A... | 399 |
5,978 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_12 | 3 | Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$ | Let $BC$ be the unknown side length. By Law of Cosines we have that $BC = \sqrt{269-260\cos{A}}$ . We notice that $\cos{A}$ should be negative to optimize $BC$ so $A$ is between $90$ and $180$ degrees. We also know that the value inside the square root is an integer $m$ , so $269-260\cos{A}$ should be an integer. We ca... | 399 |
5,979 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_15 | 1 | For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the sma... | Note that for any $x_i$ , for any prime $p$ $p^2 \nmid x_i$ . This provides motivation to translate $x_i$ into a binary sequence $y_i$
Let the prime factorization of $x_i$ be written as $p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots$ , where $p_i$ is the $i$ th prime number. Then, for every $p_{a_k}$ in the prime factori... | 149 |
5,980 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_15 | 3 | For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the sma... | Let $P_k$ denote the $k$ th prime.
Lemma: $x_{n+2^{k-1}} = P_k \cdot x_{n}$ for all $0 \leq n \leq 2^{k-1} - 1.$
$\mathbf{\mathrm{Proof:}}$
We can prove this using induction. Assume that $x_{2^{k-1}-1} = \prod_{j=1}^{k-1} P_j.$ Then, using the given recursion $x_{k+1} = \frac{x_np(x_n)}{X(x_n)}$ , we would “start fresh... | 149 |
5,981 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_1 | 1 | The AIME Triathlon consists of a half-mile swim, a 30-mile bicycle ride, and an eight-mile run. Tom swims, bicycles, and runs at constant rates. He runs fives times as fast as he swims, and he bicycles twice as fast as he runs. Tom completes the AIME Triathlon in four and a quarter hours. How many minutes does he spend... | Let $r$ represent the rate Tom swims in miles per minute. Then we have
$\frac{1/2}{r} + \frac{8}{5r} + \frac{30}{10r} = 255$
Solving for $r$ , we find $r = 1/50$ , so the time Tom spends biking is $\frac{30}{(10)(1/50)} = \boxed{150}$ minutes. | 150 |
5,982 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_2 | 1 | Find the number of five-digit positive integers, $n$ , that satisfy the following conditions: | The number takes a form of $5\text{x,y,z}5$ , in which $5|(x+y+z)$ . Let $x$ and $y$ be arbitrary digits. For each pair of $x,y$ , there are exactly two values of $z$ that satisfy the condition of $5|(x+y+z)$ . Therefore, the answer is $10\times10\times2=\boxed{200}$ | 200 |
5,983 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_4 | 1 | In the array of 13 squares shown below, 8 squares are colored red, and the remaining 5 squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated 90 degrees around the central square is $\frac{1}{n}$ , where is a posit... | When the array appears the same after a 90-degree rotation, the top formation must look the same as the right formation, which looks the same as the bottom one, which looks the same as the right one. There are four of the same configuration. There are not enough red squares for these to be all red, nor are there enough... | 429 |
5,984 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5 | 1 | The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$ | We note that $8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3$ . Therefore, we have that $9x^3 = (x+1)^3$ , so it follows that $x\sqrt[3]{9} = x+1$ . Solving for $x$ yields $\frac{1}{\sqrt[3]{9}-1} = \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$ , so the answer is $\boxed{98}$ | 98 |
5,985 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5 | 2 | The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$ | Let $r$ be the real root of the given polynomial . Now define the cubic polynomial $Q(x)=-x^3-3x^2-3x+8$ . Note that $1/r$ must be a root of $Q$ . However we can simplify $Q$ as $Q(x)=9-(x+1)^3$ , so we must have that $(\frac{1}{r}+1)^3=9$ . Thus $\frac{1}{r}=\sqrt[3]{9}-1$ , and $r=\frac{1}{\sqrt[3]{9}-1}$ . We can th... | 98 |
5,986 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5 | 3 | The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$ | It is clear that for the algebraic degree of $x$ to be $3$ that there exists some cubefree integer $p$ and positive integers $m,n$ such that $a = m^3p$ and $b = n^3p^2$ (it is possible that $b = n^3p$ , but then the problem wouldn't ask for both an $a$ and $b$ ). Let $f_1$ be the automorphism over $\mathbb{Q}[\sqrt[3]{... | 98 |
5,987 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5 | 4 | The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$ | We have $cx-1=\sqrt[3]{a}+\sqrt[3]{b}.$ Therefore $(cx-1)^3=(\sqrt[3]{a}+\sqrt[3]{b})^3=a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})=a+b+3\sqrt[3]{ab}(cx-1).$ We have \[c^3x^3-3c^2x^2-(3c\sqrt[3]{ab}-3c)x-(a+b+1-3\sqrt[3]{ab})=0.\] We will find $a,b,c$ so that the equation is equivalent to the original one. Let $\dfrac{3... | 98 |
5,988 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_8 | 1 | The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by 1000. | We start with the same method as above. The domain of the arcsin function is $[-1, 1]$ , so $-1 \le \log_{m}(nx) \le 1$
\[\frac{1}{m} \le nx \le m\] \[\frac{1}{mn} \le x \le \frac{m}{n}\] \[\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}\] \[n = 2013m - \frac{2013}{m}\]
For $n$ to be an integer, $m$ must divide $2013$ , an... | 371 |
5,989 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_8 | 2 | The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by 1000. | Note that we need $-1\le f(x)\le 1$ , and this eventually gets to $\frac{m^2-1}{mn}=\frac{1}{2013}$ . From there, break out the quadratic formula and note that \[m= \frac{n+\sqrt{n^2+4026^2}}{2013\times 2}.\] Then we realize that the square root, call it $a$ , must be an integer. Then $(a-n)(a+n)=4026^2.$
Observe caref... | 371 |
5,990 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9 | 1 | A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positi... | Let $M$ and $N$ be the points on $\overline{AB}$ and $\overline{AC}$ , respectively, where the paper is folded. Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it. [asy] import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pai... | 113 |
5,991 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9 | 2 | A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positi... | Let $P$ and $Q$ be the points on $\overline{AB}$ and $\overline{AC}$ , respectively, where the paper is folded.
Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it.
Let $a$ $b$ , and $x$ be the lengths $AP$ $AQ$ , and $PQ$ , respectively.
We have $PD = a$ $QD = b$ $BP = 12 - a$ $CQ = 12 - b$ $BD = 9... | 113 |
5,992 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9 | 3 | A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positi... | Proceed with the same labeling as in Solution 1.
$\angle B = \angle C = \angle A = \angle PDQ = 60^\circ$
$\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CQD + \angle C = 180^\circ$
Therefore, $\angle PDB = \angle CQD$
Similarly, $\angle BPD = \angle QDC$
Now, $\bigtriangleup BPD$ and $\bigtriangleup CDQ$ a... | 113 |
5,993 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9 | 4 | A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positi... | We let the original position of $A$ be $A$ , and the position of $A$ after folding be $D$ . Also, we put the triangle on the coordinate plane such that $A=(0,0)$ $B=(-6,-6\sqrt3)$ $C=(6,-6\sqrt3)$ , and $D=(3,-6\sqrt3)$
[asy] size(10cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", ... | 113 |
5,994 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | 1 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are th... | $N$ must be some multiple of $\text{lcm}(14, 15, 16)= 2^{4}\cdot 3\cdot 5\cdot 7$ ; this $lcm$ is hereby denoted $k$ and $N = qk$
$1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $10$ , and $12$ all divide $k$ , so $x, y, z = 9, 11, 13$
We have the following three modulo equations:
$nk\equiv 3\pmod{9}$
$nk\equiv 3\pmod{11}$
$nk\equiv 3... | 148 |
5,995 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | 2 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are th... | Note that the number of play blocks is a multiple of the LCM of $16$ $15$ , and $14$ . The value of this can be found to be $(16)(15)(7) = 1680$ . This number is also divisible by $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $10$ , and $12$ , thus, the three numbers $x, y, z$ are $9, 11, 13$
Thus, $1680k\equiv 3$ when taken mod $9$... | 148 |
5,996 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | 3 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are th... | It is obvious that $N=a\cdot 2^4 \cdot 3\cdot 5\cdot 7$ and so the only mod $3$ number of students are $9, 11, 13$ . Therefore, $N=1287\cdot k+3$ . Try some approaches and you will see that this one is one of the few successful ones:
Start by setting the two $N$ equations together, then we get $1680a=1287k+3$ . Divide ... | 148 |
5,997 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | 4 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are th... | We start by noticing that $N = a\textbf{lcm}(14, 15, 16) = 1680a$ for some integer $a$ in order to satisfy the first condition.
Next, we satisfy the second condition. Since $x<y<z < 14$ must leave a remainder when dividing $1680a$ , they are not divisors of $1680x$ . Thus, we can eliminate all $y \le 14$ s.t. $\gcd(y, ... | 148 |
5,998 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | 5 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are th... | As the problem stated, the number of boxes is definitely a multiple of $lcm(14,15,16)=1680$ , so we assume total number of boxes is $1680k$
Then, according to $(b)$ statement, we get $1680k \equiv 3 \pmod x \equiv 3 \pmod {y} \equiv 3 \pmod {z}$ . So we have $lcm(x,y,z)+3+m\cdot lcm(x,y,z)=1680k$ , we just write it to ... | 148 |
5,999 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | 6 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are th... | From part (a), we know that $2^4\cdot3\cdot5\cdot7 | N$ . From part (b), we know that $N \equiv 3 \pmod {1287}$ . We can expand on part (a) by saying that $N = 1680k$ for some $k$ . Rather than taking the three modulos together, we take them individually. \[1680k \equiv 6k \equiv 3 \pmod 9\] \[k \equiv 2^{-1} \pmod 9\]... | 148 |
6,000 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_13 | 1 | Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{... | Note that every $B_nC_n$ is parallel to each other for any nonnegative $n$ . Also, the area we seek is simply the ratio $k=\frac{[B_0B_1C_1]}{[B_0B_1C_1]+[C_1C_0B_0]}$ , because it repeats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90.
For ease, all ratios I will use to sol... | 961 |
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