problem stringlengths 10 5.15k | answer stringlengths 0 1.22k | solution stringlengths 0 11.1k | difficulty float64 0.75 2.02k | difficulty_raw listlengths 3 8 |
|---|---|---|---|---|
A right triangle has area 5 and a hypotenuse of length 5. Find its perimeter. | 5+3 \sqrt{5} | If $x, y$ denote the legs, then $x y=10$ and $x^{2}+y^{2}=25$, so $x+y+\sqrt{x^{2}+y^{2}}=\sqrt{\left(x^{2}+y^{2}\right)+2 x y}+5=\sqrt{45}+5=5+3 \sqrt{5}$. | 4 | [
4,
4,
4,
4,
4,
4,
4,
4
] |
A bar of chocolate is made of 10 distinguishable triangles as shown below. How many ways are there to divide the bar, along the edges of the triangles, into two or more contiguous pieces? | 1689 | Every way to divide the bar can be described as a nonempty set of edges to break, with the condition that every endpoint of a broken edge is either on the boundary of the bar or connects to another broken edge. Let the center edge have endpoints $X$ and $Y$. We do casework on whether the center edge is broken. If the c... | 6 | [
6,
7,
6,
6,
5,
7,
5,
6
] |
Suppose that $m$ and $n$ are integers with $1 \leq m \leq 49$ and $n \geq 0$ such that $m$ divides $n^{n+1}+1$. What is the number of possible values of $m$ ? | 29 | If $n$ is even, $n+1 \mid n^{n+1}+1$, so we can cover all odd $m$. If $m$ is even and $m \mid n^{n+1}+1$, then $n$ must be odd, so $n+1$ is even, and $m$ cannot be divisible by 4 or any prime congruent to $3(\bmod 4)$. Conversely, if $m / 2$ has all factors $1(\bmod 4)$, then by CRT there exists $N \equiv 1(\bmod 4)$ s... | 5.5 | [
5,
6,
6,
5,
5,
6,
5,
6
] |
Let $n$ be the answer to this problem. Find the minimum number of colors needed to color the divisors of $(n-24)$! such that no two distinct divisors $s, t$ of the same color satisfy $s \mid t$. | 50 | We first answer the following question. Find the minimum number of colors needed to color the divisors of $m$ such that no two distinct divisors $s, t$ of the same color satisfy $s \mid t$. Prime factorize $m=p_{1}^{e_{1}} \ldots p_{k}^{e_{k}}$. Note that the elements $$\begin{aligned} & 1, p_{1}, p_{1}^{2}, \ldots, p_... | 6.75 | [
7,
7,
7,
7,
6,
7,
7,
6
] |
Consider sequences \(a\) of the form \(a=\left(a_{1}, a_{2}, \ldots, a_{20}\right)\) such that each term \(a_{i}\) is either 0 or 1. For each such sequence \(a\), we can produce a sequence \(b=\left(b_{1}, b_{2}, \ldots, b_{20}\right)\), where \(b_{i}= \begin{cases}a_{i}+a_{i+1} & i=1 \\ a_{i-1}+a_{i}+a_{i+1} & 1<i<20 ... | 64 | Let the two sequences be \(b\) and \(\hat{b}\). Then, observe that given \(a\), if \(b_{1}=\hat{b}_{1}\) and \(b_{2}=\hat{b}_{2}\), then \(b=\hat{b}\) (since \(a\) will uniquely determine the remaining elements in \(b\) and \(\hat{b}\)). Thus, \(b\) and \(\hat{b}\) must start with \((1,0, \ldots)\) and \((0,1, \ldots)\... | 5.75 | [
7,
6,
6,
6,
5,
4,
5,
7
] |
An entry in a grid is called a saddle point if it is the largest number in its row and the smallest number in its column. Suppose that each cell in a $3 \times 3$ grid is filled with a real number, each chosen independently and uniformly at random from the interval $[0,1]$. Compute the probability that this grid has at... | \frac{3}{10} | With probability 1, all entries of the matrix are unique. If this is the case, we claim there can only be one saddle point. To see this, suppose $A_{i j}$ and $A_{k l}$ are both saddle points. They cannot be in the same row, since they cannot both be the greatest number in the same row, and similarly they cannot be in ... | 5.875 | [
6,
6,
6,
6,
6,
6,
6,
5
] |
Compute all ordered triples $(x, y, z)$ of real numbers satisfying the following system of equations: $$\begin{aligned} x y+z & =40 \\ x z+y & =51 \\ x+y+z & =19 \end{aligned}$$ | (12,3,4),(6,5.4,7.6) | Solution 1: By adding the first two equations, we can get $$x y+z+x z+y=(x+1)(y+z)=91$$ From the third equation we have $$(x+1)+(y+z)=19+1=20$$ so $x+1$ and $y+z$ are the two roots of $t^{2}-20 t+91=0$ by Vieta's theorem. As the quadratic equation can be decomposed into $$(t-7)(t-13)=0$$ we know that either $x=6, y+z=1... | 5.5 | [
5,
5,
6,
5,
6,
6,
6,
5
] |
In $\triangle A B C, A B=2019, B C=2020$, and $C A=2021$. Yannick draws three regular $n$-gons in the plane of $\triangle A B C$ so that each $n$-gon shares a side with a distinct side of $\triangle A B C$ and no two of the $n$-gons overlap. What is the maximum possible value of $n$? | 11 | If any $n$-gon is drawn on the same side of one side of $\triangle A B C$ as $\triangle A B C$ itself, it will necessarily overlap with another triangle whenever $n>3$. Thus either $n=3$ or the triangles are all outside $A B C$. The interior angle of a regular $n$-gon is $180^{\circ} \cdot \frac{n-2}{n}$, so we require... | 6 | [
6,
6,
6,
6,
7,
6,
6,
5
] |
Find the number of subsets $S$ of $\{1,2, \ldots 6\}$ satisfying the following conditions: - $S$ is non-empty. - No subset of $S$ has the property that the sum of its elements is 10. | 34 | We do casework based on the largest element of $S$. Call a set $n$-free if none of its subsets have elements summing to $n$. Case 1: The largest element of $S$ is 6. Then $4 \notin S$. If $5 \notin S$, then we wish to find all 4-free subsets of $\{1,2,3\}$ (note that $1+2+3=6<10$). We just cannot include both 1,3, so w... | 6.125 | [
6,
6,
6,
7,
5,
6,
7,
6
] |
Given that $62^{2}+122^{2}=18728$, find positive integers $(n, m)$ such that $n^{2}+m^{2}=9364$. | (30,92) \text{ OR } (92,30) | If $a^{2}+b^{2}=2 c$, then $\left(\frac{a+b}{2}\right)^{2}+\left(\frac{a-b}{2}\right)^{2}=\frac{2 a^{2}+2 b^{2}}{4}=\frac{a^{2}+b^{2}}{2}=c$. Thus, $n=\frac{62+122}{2}=92$ and $m=\frac{122-62}{2}=30$ works. | 3.875 | [
3,
3,
4,
4,
5,
4,
4,
4
] |
There are 21 competitors with distinct skill levels numbered $1,2, \ldots, 21$. They participate in a pingpong tournament as follows. First, a random competitor is chosen to be "active", while the rest are "inactive." Every round, a random inactive competitor is chosen to play against the current active one. The player... | \frac{47}{42} | Solution 1: Insert a player with skill level 0, who will be the first active player (and lose their first game). If Alice plays after any of the players with skill level $12,13, \ldots, 21$, which happens with probability $\frac{10}{11}$, then she will play exactly 1 game. If Alice is the first of the players with skil... | 7.375 | [
7,
8,
7,
7,
7,
8,
7,
8
] |
Let $A B C$ be a triangle and $D$ a point on $B C$ such that $A B=\sqrt{2}, A C=\sqrt{3}, \angle B A D=30^{\circ}$, and $\angle C A D=45^{\circ}$. Find $A D$. | \frac{\sqrt{6}}{2} | Note that $[B A D]+[C A D]=[A B C]$. If $\alpha_{1}=\angle B A D, \alpha_{2}=\angle C A D$, then we deduce $\frac{\sin \left(\alpha_{1}+\alpha_{2}\right)}{A D}=\frac{\sin \alpha_{1}}{A C}+\frac{\sin \alpha_{2}}{A B}$ upon division by $A B \cdot A C \cdot A D$. Now $$A D=\frac{\sin \left(30^{\circ}+45^{\circ}\right)}{\f... | 6 | [
6,
6,
6,
6,
7,
5,
6,
6
] |
Let $n$ be the answer to this problem. Suppose square $ABCD$ has side-length 3. Then, congruent non-overlapping squares $EHGF$ and $IHJK$ of side-length $\frac{n}{6}$ are drawn such that $A, C$, and $H$ are collinear, $E$ lies on $BC$ and $I$ lies on $CD$. Given that $AJG$ is an equilateral triangle, then the area of $... | 48 | The fact that $EHGF$ and $IHJK$ have side length $n/6$ ends up being irrelevant. Since $A$ and $H$ are both equidistant from $G$ and $J$, we conclude that the line $ACHM$ is the perpendicular bisector of $GJ$. Now, define the point $C^{\prime}$ so that the spiral similarity centered at $J$ sends $M$ and $H$ to $C^{\pri... | 7.125 | [
7,
7,
7,
7,
7,
7,
7,
8
] |
A bug is on one exterior vertex of solid $S$, a $3 \times 3 \times 3$ cube that has its center $1 \times 1 \times 1$ cube removed, and wishes to travel to the opposite exterior vertex. Let $O$ denote the outer surface of $S$ (formed by the surface of the $3 \times 3 \times 3$ cube). Let $L(S)$ denote the length of the ... | \frac{\sqrt{29}}{3 \sqrt{5}} \text{ OR } \frac{\sqrt{145}}{15} | By $\left(^{*}\right)$, the shortest route in $O$ has length $2 \sqrt{1.5^{2}+3^{2}}=3 \sqrt{5}$. By $\left({ }^{* *}\right)$, the shortest route overall (in $S$ ) has length $2 \sqrt{1.5^{2}+1^{2}+2^{2}}=\sqrt{3^{2}+2^{2}+4^{2}}=\sqrt{29}$. Therefore the desired ratio is $\frac{\sqrt{29}}{3 \sqrt{5}}=\frac{\sqrt{145}}... | 5.875 | [
6,
6,
6,
6,
5,
6,
6,
6
] |
Consider a $9 \times 9$ grid of squares. Haruki fills each square in this grid with an integer between 1 and 9 , inclusive. The grid is called a super-sudoku if each of the following three conditions hold: - Each column in the grid contains each of the numbers $1,2,3,4,5,6,7,8,9$ exactly once. - Each row in the grid co... | 0 | Without loss of generality, suppose that the top left corner contains a 1 , and examine the top left $3 \times 4$ : \begin{tabular}{|c|c|c|c|} \hline 1 & x & x & x \\ \hline x & x & x & $*$ \\ \hline x & x & x & $*$ \\ \hline \end{tabular} There cannot be another 1 in any of the cells marked with an x , but the $3 \tim... | 7.5 | [
7,
6,
9,
8,
7,
9,
7,
7
] |
On each side of a 6 by 8 rectangle, construct an equilateral triangle with that side as one edge such that the interior of the triangle intersects the interior of the rectangle. What is the total area of all regions that are contained in exactly 3 of the 4 equilateral triangles? | \frac{96 \sqrt{3}-154}{\sqrt{3}} \text{ OR } \frac{288-154 \sqrt{3}}{3} \text{ OR } 96-\frac{154}{\sqrt{3}} \text{ OR } 96-\frac{154 \sqrt{3}}{3} | Let the rectangle be $A B C D$ with $A B=8$ and $B C=6$. Let the four equilateral triangles be $A B P_{1}, B C P_{2}, C D P_{3}$, and $D A P_{4}$ (for convenience, call them the $P_{1^{-}}, P_{2^{-}}, P_{3^{-}}, P_{4^{-}}$triangles). Let $W=A P_{1} \cap D P_{3}, X=A P_{1} \cap D P_{4}$, and $Y=D P_{4} \cap C P_{2}$. Re... | 6.75 | [
7,
6,
7,
7,
7,
7,
7,
6
] |
Alice and Bob are playing in the forest. They have six sticks of length $1,2,3,4,5,6$ inches. Somehow, they have managed to arrange these sticks, such that they form the sides of an equiangular hexagon. Compute the sum of all possible values of the area of this hexagon. | 33 \sqrt{3} | Let the side lengths, in counterclockwise order, be $a, b, c, d, e, f$. Place the hexagon on the coordinate plane with edge $a$ parallel to the $x$-axis and the intersection between edge $a$ and edge $f$ at the origin (oriented so that edge $b$ lies in the first quadrant). If you travel along all six sides of the hexag... | 6.125 | [
6,
7,
6,
7,
5,
6,
7,
5
] |
Tessa has a unit cube, on which each vertex is labeled by a distinct integer between 1 and 8 inclusive. She also has a deck of 8 cards, 4 of which are black and 4 of which are white. At each step she draws a card from the deck, and if the card is black, she simultaneously replaces the number on each vertex by the sum o... | 42648 | The order of the deck does not matter as black cards and white cards commute, therefore we can assume that the cards are alternating black and white, and only worry about the arrangement of the numbers. After each pair of black and white cards, each number is replaced by the sum of two times the edge neighbors and thre... | 7 | [
7,
7,
7,
7,
7,
7,
7,
7
] |
There exist unique nonnegative integers $A, B$ between 0 and 9, inclusive, such that $(1001 \cdot A+110 \cdot B)^{2}=57,108,249$. Find $10 \cdot A+B$. | 75 | We only need to bound for $A B 00$; in other words, $A B^{2} \leq 5710$ but $(A B+1)^{2} \geq 5710$. A quick check gives $A B=75$. (Lots of ways to get this...) | 3.5 | [
4,
3,
4,
3,
3,
4,
4,
3
] |
It is midnight on April 29th, and Abigail is listening to a song by her favorite artist while staring at her clock, which has an hour, minute, and second hand. These hands move continuously. Between two consecutive midnights, compute the number of times the hour, minute, and second hands form two equal angles and no tw... | 5700 | Let $t \in[0,2]$ represent the position of the hour hand, i.e., how many full revolutions it has made. Then, the position of the minute hand is $12 t$ (it makes 12 full revolutions per 1 revolution of the hour hand), and the position of the second hand is $720 t$ (it makes 60 full revolutions per 1 revolution of the mi... | 7.125 | [
7,
8,
7,
7,
7,
7,
7,
7
] |
Let $N$ be the largest positive integer that can be expressed as a 2013-digit base -4 number. What is the remainder when $N$ is divided by 210? | 51 | The largest is $\sum_{i=0}^{1006} 3 \cdot 4^{2 i}=3 \frac{16^{1007}-1}{16-1}=\frac{16^{1007}-1}{5}$. This is $1(\bmod 2), 0(\bmod 3), 3 \cdot 1007 \equiv 21 \equiv 1(\bmod 5)$, and $3\left(2^{1007}-1\right) \equiv 3\left(2^{8}-1\right) \equiv 3\left(2^{2}-1\right) \equiv 2$ $(\bmod 7)$, so we need $1(\bmod 10)$ and $9(... | 5.875 | [
6,
6,
6,
6,
6,
5,
6,
6
] |
Let $f(x)=x^{3}+3 x-1$ have roots $a, b, c$. Given that $$\frac{1}{a^{3}+b^{3}}+\frac{1}{b^{3}+c^{3}}+\frac{1}{c^{3}+a^{3}}$$ can be written as $\frac{m}{n}$, where $m, n$ are positive integers and $\operatorname{gcd}(m, n)=1$, find $100 m+n$. | 3989 | We know that $a^{3}=-3 a+1$ and similarly for $b, c$, so $$\frac{1}{a^{3}+b^{3}}=\frac{1}{2-3 a-3 b}=\frac{1}{2+3 c}=\frac{1}{3(2 / 3+c)}$$ Now, $$f(x-2 / 3)=x^{3}-2 x^{2}+\frac{13}{3} x-\frac{89}{27}$$ has roots $a+2 / 3, b+2 / 3$, and $c+2 / 3$. Thus the answer is, by Vieta's formulas, $$\frac{1}{3} \frac{(a+2 / 3)(b... | 6.875 | [
7,
7,
6,
7,
7,
7,
7,
7
] |
Let $A B C D$ be a convex trapezoid such that $\angle A B C=\angle B C D=90^{\circ}, A B=3, B C=6$, and $C D=12$. Among all points $X$ inside the trapezoid satisfying $\angle X B C=\angle X D A$, compute the minimum possible value of $C X$. | \sqrt{113}-\sqrt{65} | Let $P=A D \cap B C$. Then, we see that the locus if $X$ is the arc $\widehat{B D}$ of $\odot(P B D)$. Let $O$ be its center, and $R$ be the radius of $\odot(P B D)$, then the answer is $C O-R$. Let $T$ be the second intersection of $\odot(P B D)$ and $C D$. We can compute $B P=2$, so by Power of Point, $C T \cdot C D=... | 6.875 | [
7,
7,
6,
7,
7,
7,
7,
7
] |
Let $A B C D$ be a square of side length 5. A circle passing through $A$ is tangent to segment $C D$ at $T$ and meets $A B$ and $A D$ again at $X \neq A$ and $Y \neq A$, respectively. Given that $X Y=6$, compute $A T$. | \sqrt{30} | Let $O$ be the center of the circle, and let $Z$ be the foot from $O$ to $A D$. Since $X Y$ is a diameter, $O T=Z D=3$, so $A Z=2$. Then $O Z=\sqrt{5}$ and $A T=\sqrt{O Z^{2}+25}=\sqrt{30}$. | 5.125 | [
5,
5,
5,
5,
5,
5,
6,
5
] |
The number $989 \cdot 1001 \cdot 1007+320$ can be written as the product of three distinct primes $p, q, r$ with $p<q<r$. Find $(p, q, r)$. | (991,997,1009) | Let $f(x)=x(x-12)(x+6)+320=x^{3}-6 x^{2}-72 x+320$, so that $f(1001)=989 \cdot 1001 \cdot 1007+320$. But $f(4)=4(-8)(10)+320=0$, so $f(x)=(x-4)\left(x^{2}-2 x-80\right)=(x-4)(x-10)(x+8)$ Thus $f(1001)=991 \cdot 997 \cdot 1009$, as desired. | 5.5 | [
6,
5,
6,
5,
5,
5,
6,
6
] |
If you roll four fair 6-sided dice, what is the probability that at least three of them will show the same value? | \frac{7}{72} | We have two cases: either three of the dice show one value and the last shows a different value, or all four dice show the same value. In the first case, there are six choices for the value of the dice which are the same and $\binom{4}{3}$ choice for which dice show that value. Then there are 5 choices for the last die... | 3.5 | [
3,
3,
4,
4,
3,
3,
4,
4
] |
A 50-card deck consists of 4 cards labeled " $i$ " for $i=1,2, \ldots, 12$ and 2 cards labeled " 13 ". If Bob randomly chooses 2 cards from the deck without replacement, what is the probability that his 2 cards have the same label? | \frac{73}{1225} | All pairs of distinct cards (where we distinguish cards even with the same label) are equally likely. There are $\binom{2}{2}+12\binom{4}{2}=73$ pairs of cards with the same label and $\binom{50}{2}=100 \cdot \frac{49}{4}=1225$ pairs of cards overall, so the desired probability is $\frac{73}{1225}$. | 4 | [
5,
4,
4,
3,
4,
4,
4,
4
] |
Compute the number of positive four-digit multiples of 11 whose sum of digits (in base ten) is divisible by 11. | 72 | Let an arbitrary such number be \(\overline{a b c d}\). Then, we desire \(11 \mid a+b+c+d\) and \(11 \mid a-b+c-d\), where the latter comes from the well-known divisibility trick for 11. Sums and differences of multiples of 11 must also be multiples of 11, so this is equivalent to desiring \(11 \mid a+c\) and \(11 \mid... | 4 | [
3,
4,
5,
4,
4,
4,
4,
4
] |
How many of the first 1000 positive integers can be written as the sum of finitely many distinct numbers from the sequence $3^{0}, 3^{1}, 3^{2}, \ldots$? | 105 | We want to find which integers have only 0 's and 1 's in their base 3 representation. Note that $1000_{10}=1101001_{3}$. We can construct a bijection from all such numbers to the binary strings, by mapping $x_{3} \leftrightarrow x_{2}$. Since $1101001_{2}=105_{10}$, we conclude that the answer is 105. | 4.875 | [
5,
5,
5,
4,
6,
5,
4,
5
] |
The classrooms at MIT are each identified with a positive integer (with no leading zeroes). One day, as President Reif walks down the Infinite Corridor, he notices that a digit zero on a room sign has fallen off. Let $N$ be the original number of the room, and let $M$ be the room number as shown on the sign. The smalle... | 2031 | Let $A$ represent the portion of $N$ to the right of the deleted zero, and $B$ represent the rest of $N$. For example, if the unique zero in $N=12034$ is removed, then $A=34$ and $B=12000$. Then, $\frac{M}{N}=\frac{A+B / 10}{A+B}=1-\frac{9}{10} \frac{B}{N}$. The maximum value for $B / N$ is 1 , which is achieved when $... | 5.75 | [
7,
7,
5,
5,
6,
6,
5,
5
] |
Pentagon $S P E A K$ is inscribed in triangle $N O W$ such that $S$ and $P$ lie on segment $N O, K$ and $A$ lie on segment $N W$, and $E$ lies on segment $O W$. Suppose that $N S=S P=P O$ and $N K=K A=A W$. Given that $E P=E K=5$ and $E A=E S=6$, compute $O W$. | \frac{3 \sqrt{610}}{5} | Note that $[E S K]=[E P A]$, since one has half the base but double the height. Since the sides are the same, we must have $\sin \angle S E K=\sin \angle P E A$, so $\angle S E K+\angle P E A=180^{\circ}$. Let $O W=3 x$, so $S K=x$ and $P A=2 x$. Then by the law of cosines $$\begin{aligned} x^{2} & =61-60 \cos \angle S... | 6.625 | [
7,
6,
7,
7,
7,
6,
6,
7
] |
Alice thinks of four positive integers $a \leq b \leq c \leq d$ satisfying $\{a b+c d, a c+b d, a d+b c\}=\{40,70,100\}$. What are all the possible tuples $(a, b, c, d)$ that Alice could be thinking of? | (1,4,6,16) | Since $a b \cdot c d=a c \cdot b d=a d \cdot b c$, the largest sum among $a b+c d, a c+b d, a d+b c$ will be the one with the largest difference between the two quantities, so $a b+c d=100, a c+b d=70, a d+b c=40$. Consider the sum of each pair of equations, which gives $(a+b)(c+d)=110,(a+c)(b+d)=140,(a+$ $d)(b+c)=170$... | 6.625 | [
7,
6,
7,
6,
6,
7,
7,
7
] |
How many pairs of real numbers $(x, y)$ satisfy the equation $y^{4}-y^{2}=x y^{3}-x y=x^{3} y-x y=x^{4}-x^{2}=0$? | 9 | We can see that if they solve the first and fourth equations, they are automatically solutions to the second and third equations. Hence, the solutions are just the $3^{2}=9$ points where $x, y$ can be any of $-1,0,1$. | 3.75 | [
4,
4,
4,
3,
4,
3,
4,
4
] |
A square is inscribed in a circle of radius 1. Find the perimeter of the square. | 4 \sqrt{2} | The square has diagonal length 2, so side length $\sqrt{2}$ and perimeter $4 \sqrt{2}$. | 1 | [
1,
1,
1,
1,
1,
1,
1,
1
] |
Find the smallest positive integer $n$ such that $\underbrace{2^{2 \cdot 2}}_{n}>3^{3^{3^{3}}}$. (The notation $\underbrace{2^{2^{2}}}_{n}$, is used to denote a power tower with $n 2$ 's. For example, $\underbrace{2^{22^{2}}}_{n}$ with $n=4$ would equal $2^{2^{2^{2}}}$.) | 6 | Clearly, $n \geq 5$. When we take $n=5$, we have $$2^{2^{2^{2^{2}}}}=2^{2^{16}}<3^{3^{27}}=3^{3^{3^{3}}}.$$ On the other hand, when $n=6$, we have $$2^{2^{2^{2^{2^{2}}}}}=2^{2^{65536}}=4^{2^{65535}}>4^{4^{27}}>3^{3^{27}}=3^{3^{3^{3}}}.$$ Our answer is thus $n=6$. | 6 | [
6,
6,
6,
6,
6,
6,
6,
6
] |
David has a unit triangular array of 10 points, 4 on each side. A looping path is a sequence $A_{1}, A_{2}, \ldots, A_{10}$ containing each of the 10 points exactly once, such that $A_{i}$ and $A_{i+1}$ are adjacent (exactly 1 unit apart) for $i=1,2, \ldots, 10$. (Here $A_{11}=A_{1}$.) Find the number of looping paths ... | 60 | There are $10 \cdot 2$ times as many loop sequences as loops. To count the number of loops, first focus on the three corners of the array: their edges are uniquely determined. It's now easy to see there are 3 loops (they form " $V$-shapes"), so the answer is $10 \cdot 2 \cdot 3=60$. | 5.125 | [
6,
5,
5,
5,
4,
5,
6,
5
] |
In the game of rock-paper-scissors-lizard-Spock, rock defeats scissors and lizard, paper defeats rock and Spock, scissors defeats paper and lizard, lizard defeats paper and Spock, and Spock defeats rock and scissors. If three people each play a game of rock-paper-scissors-lizard-Spock at the same time by choosing one o... | \frac{12}{25} | Let the three players be $A, B, C$. Our answer will simply be the sum of the probability that $A$ beats both $B$ and $C$, the probability that $B$ beats both $C$ and $A$, and the probability that $C$ beats $A$ and $B$, because these events are all mutually exclusive. By symmetry, these three probabilities are the same,... | 4.125 | [
4,
4,
4,
5,
4,
4,
4,
4
] |
Two fair six-sided dice are rolled. What is the probability that their sum is at least 10? | \frac{1}{6} | There are $3,2,1$ outcomes with sum $10,11,12$, so the probability is $\frac{3+2+1}{6^{2}}=\frac{1}{6}$. | 2 | [
2,
2,
2,
2,
2,
2,
2,
2
] |
Points $G$ and $N$ are chosen on the interiors of sides $E D$ and $D O$ of unit square $D O M E$, so that pentagon GNOME has only two distinct side lengths. The sum of all possible areas of quadrilateral $N O M E$ can be expressed as $\frac{a-b \sqrt{c}}{d}$, where $a, b, c, d$ are positive integers such that $\operato... | 10324 | Since $M O=M E=1$, but $O N$ and $G E$ are both less than 1, we must have either $O N=N G=G E=x$ (call this case 1) or $O N=G E=x, N G=1$ (call this case 2). Either way, the area of $N O M E$ (a trapezoid) is $\frac{1+x}{2}$, and triangle $N G T$ is a $45-45-90$ triangle. In case 1, we have $1=O N+N T=x\left(1+\frac{\s... | 6.5 | [
6,
6,
6,
7,
7,
6,
7,
7
] |
Find the sum of $\frac{1}{n}$ over all positive integers $n$ with the property that the decimal representation of $\frac{1}{n}$ terminates. | \sqrt{\frac{5}{2}} | The decimal representation of $\frac{1}{n}$ terminates if and only if $n=2^{i} 5^{j}$ for some nonnegative integers $i, j$, so our desired sum is $$\sum_{i \geq 0} \sum_{j \geq 0} 2^{-i} 5^{-j}=\sum_{i \geq 0} 2^{-i} \sum_{j \geq 0} 5^{-j}=\left(1-2^{-1}\right)^{-1}\left(1-5^{-1}\right)^{-1}=\frac{2}{1} \frac{5}{4}=\fr... | 3.625 | [
4,
3,
4,
4,
3,
4,
3,
4
] |
A 24-hour digital clock shows times $h: m: s$, where $h, m$, and $s$ are integers with $0 \leq h \leq 23$, $0 \leq m \leq 59$, and $0 \leq s \leq 59$. How many times $h: m: s$ satisfy $h+m=s$? | 1164 | We are solving $h+m=s$ in $0 \leq s \leq 59,0 \leq m \leq 59$, and $0 \leq h \leq 23$. If $s \geq 24$, each $h$ corresponds to exactly 1 solution, so we get $24(59-23)=24(36)$ in this case. If $s \leq 23$, we want the number of nonnegative integer solutions to $h+m \leq 23$, which by lattice point counting (or balls an... | 4.75 | [
4,
5,
4,
5,
6,
5,
4,
5
] |
How many positive integers less than 100 are relatively prime to 200? | 40 | $401 \leq n<100$ is relatively prime to 200 if and only if it's relatively prime to 100 (200, 100 have the same prime factors). Thus our answer is $\phi(100)=100 \frac{1}{2} \frac{4}{5}=40$. | 3.125 | [
3,
4,
3,
3,
3,
3,
3,
3
] |
Given any positive integer, we can write the integer in base 12 and add together the digits of its base 12 representation. We perform this operation on the number $7^{6^{5^{3^{2^{1}}}}}$ repeatedly until a single base 12 digit remains. Find this digit. | 4 | For a positive integer $n$, let $s(n)$ be the sum of digits when $n$ is expressed in base 12. We claim that $s(n) \equiv n(\bmod 11)$ for all positive integers $n$. Indeed, if $n=d_{k} 12^{k}+d_{k-1} 12^{k-1}+\cdots+d_{0}$ with each $d_{i}$ an integer between 0 and 11, inclusive, because $12 \equiv 1(\bmod 11)$, reduci... | 7.125 | [
7,
7,
8,
7,
7,
6,
7,
8
] |
Find the number of strictly increasing sequences of nonnegative integers with the following properties: - The first term is 0 and the last term is 12. In particular, the sequence has at least two terms. - Among any two consecutive terms, exactly one of them is even. | 144 | For a natural number $n$, let $A_{n}$ be a set containing all sequences which satisfy the problem conditions but which 12 is replaced by $n$. Also, let $a_{n}$ be the size of $A_{n}$. We first consider $a_{1}$ and $a_{2}$. We get $a_{1}=1$, as the only sequence satisfying the problem conditions is 0,1. We also get $a_{... | 6.25 | [
6,
6,
7,
6,
6,
6,
6,
7
] |
Find the number of positive integers less than 1000000 that are divisible by some perfect cube greater than 1. | 168089 | Using the following code, we get the answer (denoted by the variable ans): ans $=0$ for $n$ in xrange $(1,1000000)$ : ``` divisible_by_cube = True for i in xrange(2,101): if n%(i*i*i)==0: divisible_by_cube = False break if divisible_by_cube: ans = ans + 1 print ans ``` This gives the output 168089 Alternatively, let $N... | 6 | [
5,
6,
6,
7,
7,
6,
6,
5
] |
Ben "One Hunna Dolla" Franklin is flying a kite KITE such that $I E$ is the perpendicular bisector of $K T$. Let $I E$ meet $K T$ at $R$. The midpoints of $K I, I T, T E, E K$ are $A, N, M, D$, respectively. Given that $[M A K E]=18, I T=10,[R A I N]=4$, find $[D I M E]$. | 16 | Let $[K I R]=[R I T]=a$ and $[K E R]=[T E R]=b$. We will relate all areas to $a$ and $b$. First, $$ [R A I N]=[R A I]+[I N R]=\frac{1}{2} a+\frac{1}{2} a=a $$ Next, we break up $[M A K E]=[M A D]+[A K D]+[D E M]$. We have $$ \begin{aligned} & {[M A D]=\frac{A D \cdot D M}{2}=\frac{1}{2} \cdot \frac{I E}{2} \cdot \frac{... | 6.875 | [
7,
6,
7,
7,
7,
7,
7,
7
] |
All positive integers whose binary representations (excluding leading zeroes) have at least as many 1's as 0's are put in increasing order. Compute the number of digits in the binary representation of the 200th number. | 9 | We do a rough estimation. There are 255 positive integers with at most 8 digits and a majority of them, but not more than 200, satisfy the property. Meanwhile, there are 511 positive integers with at most 9 digits, and a majority of them satisfy this property. Thus, the answer must be greater than 8 and at most 9. | 5 | [
4,
5,
6,
5,
5,
5,
5,
5
] |
Sean enters a classroom in the Memorial Hall and sees a 1 followed by 2020 0's on the blackboard. As he is early for class, he decides to go through the digits from right to left and independently erase the $n$th digit from the left with probability $\frac{n-1}{n}$. (In particular, the 1 is never erased.) Compute the e... | 681751 | Suppose Sean instead follows this equivalent procedure: he starts with $M=10 \ldots 0$, on the board, as before. Instead of erasing digits, he starts writing a new number on the board. He goes through the digits of $M$ one by one from left to right, and independently copies the $n$th digit from the left with probabilit... | 7.25 | [
7,
7,
8,
8,
7,
7,
7,
7
] |
Suppose that $a$ and $b$ are real numbers such that the line $y=a x+b$ intersects the graph of $y=x^{2}$ at two distinct points $A$ and $B$. If the coordinates of the midpoint of $A B$ are $(5,101)$, compute $a+b$. | 61 | Solution 1: Let $A=\left(r, r^{2}\right)$ and $B=\left(s, s^{2}\right)$. Since $r$ and $s$ are roots of $x^{2}-a x-b$ with midpoint 5, $r+s=10=a$ (where the last equality follows by Vieta's formula). Now, as $-r s=b$ (Vieta's formula), observe that $$202=r^{2}+s^{2}=(r+s)^{2}-2 r s=100+2 b$$ This means $b=51$, so the a... | 5 | [
5,
5,
4,
5,
5,
5,
6,
5
] |
Let $S$ be a subset of $\{1,2,3, \ldots, 12\}$ such that it is impossible to partition $S$ into $k$ disjoint subsets, each of whose elements sum to the same value, for any integer $k \geq 2$. Find the maximum possible sum of the elements of $S$. | 77 | We note that the maximum possible sum is 78 (the entire set). However, this could be partitioned into 2 subsets with sum 39: $\{1,2,3,10,11,12\}$ and $\{4,5,6,7,8,9\}$. The next largest possible sum is 77 (the entire set except 1). If $k \geq 2$ subsets each had equal sum, then they would have to be 7 subsets with sum ... | 5.625 | [
5,
6,
6,
5,
4,
6,
7,
6
] |
Betty has a $3 \times 4$ grid of dots. She colors each dot either red or maroon. Compute the number of ways Betty can color the grid such that there is no rectangle whose sides are parallel to the grid lines and whose vertices all have the same color. | 408 | First suppose no 3 by 1 row is all red or all blue. Then each row is either two red and one blue, or two blue and one red. There are 6 possible configurations of such a row, and as long as no row is repeated, there's no monochromatic rectangle This gives $6 \cdot 5 \cdot 4 \cdot 3=360$ possibilities. Now suppose we hav... | 5.75 | [
6,
6,
6,
5,
6,
6,
6,
5
] |
A number is chosen uniformly at random from the set of all positive integers with at least two digits, none of which are repeated. Find the probability that the number is even. | \frac{41}{81} | Since the number has at least two digits, all possible combinations of first and last digits have the same number of possibilities, which is \(\sum_{i=0}^{8} \frac{8!}{i!}\). Since the first digit cannot be zero, all of the last digits have 8 possible first digits, except for 0, which has 9 possible first digits. There... | 4.875 | [
5,
4,
5,
5,
5,
5,
5,
5
] |
Let \(P(x)\) be a quadratic polynomial with real coefficients. Suppose that \(P(1)=20, P(-1)=22\), and \(P(P(0))=400\). Compute the largest possible value of \(P(10)\). | 2486 | Let \(P(x)=ax^{2}+bx+c\). The given equations give us: \(a+b+c=20\) and \(a-b+c=22\). Hence \(b=-1, a+c=21\), and so the final equation gives us \(ac^{2}=400\). Substituting \(a=21-c\) and solving the cubic in \(c\), we get \(c=-4,5,20\). Of these, the smallest value \(c=-4\) (and hence \(P(x)=25x^{2}-x-4\)) ends up gi... | 5.75 | [
6,
5,
6,
6,
6,
5,
6,
6
] |
Let \(ABC\) be a triangle with \(AB=2021, AC=2022\), and \(BC=2023\). Compute the minimum value of \(AP+2BP+3CP\) over all points \(P\) in the plane. | 6068 | The minimizing point is when \(P=C\). To prove this, consider placing \(P\) at any other point \(O \neq C\). Then, by moving \(P\) from \(O\) to \(C\), the expression changes by \((AC-AO)+2(BC-BO)+3(CC-CO)<OC+2OC-3OC=0\) by the triangle inequality. Since this is negative, \(P=C\) must be the optimal point. The answer i... | 5.25 | [
5,
5,
5,
5,
6,
6,
5,
5
] |
Brave NiuNiu (a milk drink company) organizes a promotion during the Chinese New Year: one gets a red packet when buying a carton of milk of their brand, and there is one of the following characters in the red packet "虎"(Tiger), "生"(Gain), "威"(Strength). If one collects two "虎", one "生" and one "威", then they form a Ch... | 7 \frac{1}{3} | The answer is B. We can use Poisson process to get the explicit formula for the general case. Suppose that there are in total $n$ characters. The probability for the character $i$ is $p_{i}$, and we aim to collect $k_{i}$ copies of the character $i$. We denote by $N$ the first time to realize our collection, and we nee... | 7 | [
7,
7,
7,
7,
7,
7,
7,
7
] |
Farmer James invents a new currency, such that for every positive integer $n \leq 6$, there exists an $n$-coin worth $n$ ! cents. Furthermore, he has exactly $n$ copies of each $n$-coin. An integer $k$ is said to be nice if Farmer James can make $k$ cents using at least one copy of each type of coin. How many positive ... | 210 | We use the factorial base, where we denote $$ \left(d_{n} \ldots d_{1}\right)_{*}=d_{n} \times n!+\cdots+d_{1} \times 1! $$ The representation of $2018_{10}$ is $244002_{*}$ and the representation of $720_{10}$ is $100000_{*}$. The largest nice number less than $244002_{*}$ is $243321_{*}$. Notice that for the digit $d... | 6.75 | [
7,
7,
6,
6,
7,
7,
7,
7
] |
An up-right path between two lattice points $P$ and $Q$ is a path from $P$ to $Q$ that takes steps of 1 unit either up or to the right. A lattice point $(x, y)$ with $0 \leq x, y \leq 5$ is chosen uniformly at random. Compute the expected number of up-right paths from $(0,0)$ to $(5,5)$ not passing through $(x, y)$ | 175 | For a lattice point $(x, y)$, let $F(x, y)$ denote the number of up-right paths from $(0,0)$ to $(5,5)$ that don't pass through $(x, y)$, and let $$S=\sum_{0 \leq x \leq 5} \sum_{0 \leq y \leq 5} F(x, y)$$ Our answer is $\frac{S}{36}$, as there are 36 lattice points $(x, y)$ with $0 \leq x, y \leq 5$. Notice that the n... | 6.125 | [
6,
6,
6,
6,
7,
6,
6,
6
] |
Consider the paths from \((0,0)\) to \((6,3)\) that only take steps of unit length up and right. Compute the sum of the areas bounded by the path, the \(x\)-axis, and the line \(x=6\) over all such paths. | 756 | We see that the sum of the areas under the path is equal the sum of the areas above the path. Thus, the sum of the areas under the path is half the area of the rectangle times the number of paths, which is \(\frac{18\binom{9}{3}}{2}=756\). | 5.5 | [
5,
6,
5,
6,
5,
7,
5,
5
] |
Isosceles trapezoid \(ABCD\) with bases \(AB\) and \(CD\) has a point \(P\) on \(AB\) with \(AP=11, BP=27\), \(CD=34\), and \(\angle CPD=90^{\circ}\). Compute the height of isosceles trapezoid \(ABCD\). | 15 | Drop projections of \(A, P, B\) onto \(CD\) to get \(A', P', B'\). Since \(A'B'=38\) and \(CD=34\), we get that \(DA'=CB'=2\). Thus, \(P'D=9\) and \(P'C=25\). Hence, the answer is \(PP'=\sqrt{P'D \cdot P'C}=15\). | 5 | [
5,
6,
4,
5,
5,
5,
5,
5
] |
How many ways are there to place 31 knights in the cells of an $8 \times 8$ unit grid so that no two attack one another? | 68 | Consider coloring the squares of the chessboard so that 32 are black and 32 are white, and no two squares of the same color share a side. Then a knight in a square of one color only attacks squares of the opposite color. Any arrangement of knights in which all 31 are placed on the same color therefore works: there are ... | 5.75 | [
7,
6,
5,
5,
6,
4,
6,
7
] |
After viewing the John Harvard statue, a group of tourists decides to estimate the distances of nearby locations on a map by drawing a circle, centered at the statue, of radius $\sqrt{n}$ inches for each integer $2020 \leq n \leq 10000$, so that they draw 7981 circles altogether. Given that, on the map, the Johnston Ga... | 49 | Consider a coordinate system on any line $\ell$ where 0 is placed at the foot from $(0,0)$ to $\ell$. Then, by the Pythagorean theorem, a point $(x, y)$ on $\ell$ is assigned a coordinate $u$ for which $x^{2}+y^{2}=u^{2}+a$ for some fixed $a$ (dependent only on $\ell$ ). Consider this assignment of coordinates for our ... | 6.375 | [
6,
7,
6,
6,
7,
6,
6,
7
] |
Let $n$ be the answer to this problem. $a$ and $b$ are positive integers satisfying $$\begin{aligned} & 3a+5b \equiv 19 \quad(\bmod n+1) \\ & 4a+2b \equiv 25 \quad(\bmod n+1) \end{aligned}$$ Find $2a+6b$. | 96 | Let $m=n+1$, so that the conditions become $$\begin{align*} & 3a+5b \equiv 19 \quad(\bmod m) \tag{1}\\ & 4a+2b \equiv 25 \quad(\bmod m) \tag{2}\\ & 2a+6b \equiv-1 \quad(\bmod m) \tag{3} \end{align*}$$ We can subtract (2) from twice (3) to obtain $$10b \equiv-27 \quad(\bmod m)$$ Multiplying (1) by 2 and replacing $10b... | 5.625 | [
6,
5,
6,
5,
6,
6,
6,
5
] |
The elevator buttons in Harvard's Science Center form a $3 \times 2$ grid of identical buttons, and each button lights up when pressed. One day, a student is in the elevator when all the other lights in the elevator malfunction, so that only the buttons which are lit can be seen, but one cannot see which floors they co... | 44 | We first note that there are $2^{6}-1=63$ possibilities for lights in total. We now count the number of duplicates we need to subtract by casework on the number of buttons lit. To do this, we do casework on the size of the minimal "bounding box" of the lights: - If the bounding box is $1 \times 1$, the only arrangement... | 5.125 | [
4,
6,
4,
5,
5,
5,
6,
6
] |
Find the number of ordered triples of integers $(a, b, c)$ with $1 \leq a, b, c \leq 100$ and $a^{2} b+b^{2} c+c^{2} a=a b^{2}+b c^{2}+c a^{2}$ | 29800 | This factors as $(a-b)(b-c)(c-a)=0$. By the inclusion-exclusion principle, we get $3 \cdot 100^{2}-3 \cdot 100+100=29800$. | 5.25 | [
5,
5,
5,
6,
6,
6,
5,
4
] |
Lil Wayne, the rain god, determines the weather. If Lil Wayne makes it rain on any given day, the probability that he makes it rain the next day is $75 \%$. If Lil Wayne doesn't make it rain on one day, the probability that he makes it rain the next day is $25 \%$. He decides not to make it rain today. Find the smalles... | 9 | Let $p_{n}$ denote the probability that Lil Wayne makes it rain $n$ days from today. We have $p_{0}=0$ and $$ p_{n+1}=\frac{3}{4} p_{n}+\frac{1}{4}\left(1-p_{n}\right)=\frac{1}{4}+\frac{1}{2} p_{n} $$ This can be written as $$ p_{n+1}-\frac{1}{2}=\frac{1}{2}\left(p_{n}-\frac{1}{2}\right) $$ and we can check that the so... | 5.5 | [
6,
4,
5,
6,
6,
6,
6,
5
] |
Let \(ABC\) be a triangle with \(AB=13, BC=14\), and \(CA=15\). Pick points \(Q\) and \(R\) on \(AC\) and \(AB\) such that \(\angle CBQ=\angle BCR=90^{\circ}\). There exist two points \(P_{1} \neq P_{2}\) in the plane of \(ABC\) such that \(\triangle P_{1}QR, \triangle P_{2}QR\), and \(\triangle ABC\) are similar (with... | 48 | Let \(T\) be the foot of the \(A\)-altitude of \(ABC\). Recall that \(BT=5\) and \(CT=9\). Let \(T'\) be the foot of the \(P\)-altitude of \(PQR\). Since \(T'\) is the midpoint of the possibilities for \(P\), the answer is \(\sum_{P} d(P, BC)=2 d(T', BC)\). Since \(T'\) splits \(QR\) in a \(5:9\) ratio, we have \(d(T',... | 6.75 | [
6,
6,
7,
7,
7,
7,
7,
7
] |
Suppose point \(P\) is inside triangle \(ABC\). Let \(AP, BP\), and \(CP\) intersect sides \(BC, CA\), and \(AB\) at points \(D, E\), and \(F\), respectively. Suppose \(\angle APB=\angle BPC=\angle CPA, PD=\frac{1}{4}, PE=\frac{1}{5}\), and \(PF=\frac{1}{7}\). Compute \(AP+BP+CP\). | \frac{19}{12} | The key is the following lemma: Lemma: If \(\angle X=120^{\circ}\) in \(\triangle XYZ\), and the bisector of \(X\) intersects \(YZ\) at \(T\), then \(\frac{1}{XY}+\frac{1}{XZ}=\frac{1}{XT}\). Proof of the Lemma. Construct point \(W\) on \(XY\) such that \(\triangle XWT\) is equilateral. We also have \(TW \parallel XZ\)... | 6.25 | [
6,
6,
6,
6,
6,
6,
7,
7
] |
A regular tetrahedron has a square shadow of area 16 when projected onto a flat surface (light is shone perpendicular onto the plane). Compute the sidelength of the regular tetrahedron. | 4 \sqrt{2} | Imagine the shadow of the skeleton of the tetrahedron (i.e. make the entire tetrahedron translucent except for the edges). The diagonals of the square shadow must correspond to a pair of opposite edges of the tetrahedron. Both of these edges must be parallel to the plane - if they weren't, then edges corresponding to t... | 3.875 | [
4,
3,
4,
4,
4,
4,
4,
4
] |
Real numbers \(x\) and \(y\) satisfy the following equations: \(x=\log_{10}(10^{y-1}+1)-1\) and \(y=\log_{10}(10^{x}+1)-1\). Compute \(10^{x-y}\). | \frac{101}{110} | Taking 10 to the power of both sides in each equation, these equations become: \(10^{x}=\left(10^{y-1}+1\right) \cdot 10^{-1}\) and \(10^{y}=\left(10^{x}+1\right) \cdot 10^{-1}\). Let \(a=10^{x}\) and \(b=10^{y}\). Our equations become: \(10a=b/10+1\) and \(10b=a+1\) and we are asked to compute \(a/b\). Subtracting the... | 5 | [
5,
5,
6,
5,
4,
5,
6,
4
] |
The rightmost nonzero digit in the decimal expansion of 101 ! is the same as the rightmost nonzero digit of $n$ !, where $n$ is an integer greater than 101. Find the smallest possible value of $n$. | 103 | 101! has more factors of 2 than 5, so its rightmost nonzero digit is one of $2,4,6,8$. Notice that if the rightmost nonzero digit of 101 ! is $2 k(1 \leq k \leq 4)$, then 102 ! has rightmost nonzero digit $102(2 k) \equiv 4 k(\bmod 10)$, and 103 ! has rightmost nonzero digit $103(4 k) \equiv 2 k(\bmod 10)$. Hence $n=10... | 6.125 | [
7,
7,
6,
5,
6,
6,
6,
6
] |
How many different numbers are obtainable from five 5s by first concatenating some of the 5s, then multiplying them together? For example, we could do $5 \cdot 55 \cdot 55,555 \cdot 55$, or 55555, but not $5 \cdot 5$ or 2525. | 7 | If we do 55555, then we're done. Note that $5,55,555$, and 5555 all have completely distinguishable prime factorizations. This means that if we are given a product of them, we can obtain the individual terms. The number of 5555's is the exponent of 101, the number of 555's is the exponent of 37, the number of 55's is t... | 3.75 | [
3,
3,
4,
4,
4,
5,
3,
4
] |
You have a length of string and 7 beads in the 7 colors of the rainbow. You place the beads on the string as follows - you randomly pick a bead that you haven't used yet, then randomly add it to either the left end or the right end of the string. What is the probability that, at the end, the colors of the beads are the... | \frac{1}{5040} | The threading method does not depend on the colors of the beads, so at the end all configurations are equally likely. Since there are $7!=5040$ configurations in total, the probability of any particular configuration is $\frac{1}{5040}$. | 3.625 | [
4,
3,
3,
3,
4,
5,
3,
4
] |
Suppose that $x, y$, and $z$ are non-negative real numbers such that $x+y+z=1$. What is the maximum possible value of $x+y^{2}+z^{3}$ ? | 1 | Since $0 \leq y, z \leq 1$, we have $y^{2} \leq y$ and $z^{3} \leq z$. Therefore $x+y^{2}+z^{3} \leq x+y+z=1$. We can get $x+y^{2}+z^{3}=1$ by setting $(x, y, z)=(1,0,0)$. | 3.875 | [
4,
4,
4,
4,
4,
4,
3,
4
] |
What are the last 8 digits of $$11 \times 101 \times 1001 \times 10001 \times 100001 \times 1000001 \times 111 ?$$ | 19754321 | Multiply terms in a clever order. $$\begin{aligned} 11 \cdot 101 \cdot 10001 & =11,111,111 \\ 111 \cdot 1001 \cdot 1000001 & =111,111,111,111 \end{aligned}$$ The last eight digits of $11,111,111 \cdot 111,111,111,111$ are 87654321. We then just need to compute the last 8 digits of $87654321 \cdot 100001=87654321+\ldots... | 4.25 | [
4,
4,
4,
4,
5,
5,
4,
4
] |
To set up for a Fourth of July party, David is making a string of red, white, and blue balloons. He places them according to the following rules: - No red balloon is adjacent to another red balloon. - White balloons appear in groups of exactly two, and groups of white balloons are separated by at least two non-white ba... | 99 | It is possible to achieve 99 red balloons with the arrangement $$\text { WWBBBWW } \underbrace{\text { RBBBWWRBBBWW ...RBBBWW, }}_{99 \text { RBBBWW's }}$$ which contains $99 \cdot 6+7=601$ balloons. Now assume that one can construct a chain with 98 or fewer red balloons. Then there can be 99 blocks of non-red balloons... | 6.625 | [
6,
7,
6,
7,
7,
7,
7,
6
] |
Find all prime numbers $p$ for which there exists a unique $a \in\{1,2, \ldots, p\}$ such that $a^{3}-3 a+1$ is divisible by $p$. | 3 | We show that $p=3$ is the only prime that satisfies the condition. Let $f(x)=x^{3}-3 x+1$. As preparation, let's compute the roots of $f(x)$. By Cardano's formula, it can be seen that the roots are $2 \operatorname{Re} \sqrt[3]{\frac{-1}{2}+\sqrt{\left(\frac{-1}{2}\right)^{2}-\left(\frac{-3}{3}\right)^{3}}}=2 \operator... | 7.375 | [
7,
8,
7,
7,
7,
8,
8,
7
] |
In the future, MIT has attracted so many students that its buildings have become skyscrapers. Ben and Jerry decide to go ziplining together. Ben starts at the top of the Green Building, and ziplines to the bottom of the Stata Center. After waiting $a$ seconds, Jerry starts at the top of the Stata Center, and ziplines t... | 740 | Note that due to all the 3-4-5 triangles, we find $\frac{x}{z}=\frac{z}{y}=\frac{4}{3}$, so $120=x+y=\frac{25}{12} z$. Then, $$u=\frac{5}{3} x=\frac{20}{9} z=\frac{16}{15} 120=128$$ while $$v=\frac{5}{4} y=\frac{15}{16} z=\frac{9}{20} 120=54$$ Thus $u-v=74$, implying that $a=7.4$. | 5.375 | [
5,
6,
4,
6,
6,
5,
5,
6
] |
Let $n$ be the answer to this problem. An urn contains white and black balls. There are $n$ white balls and at least two balls of each color in the urn. Two balls are randomly drawn from the urn without replacement. Find the probability, in percent, that the first ball drawn is white and the second is black. | 19 | Let the number of black balls in the urn be $k \geq 2$. Then the probability of drawing a white ball first is $\frac{n}{n+k}$, and the probability of drawing a black ball second is $\frac{k}{n+k-1}$. This gives us the equation $$\frac{nk}{(n+k)(n+k-1)}=\frac{n}{100}$$ from which we get $$(n+k)(n+k-1)=100k$$ Let $m=n+k$... | 4.5 | [
4,
5,
4,
4,
5,
5,
5,
4
] |
Marcus and four of his relatives are at a party. Each pair of the five people are either friends or enemies. For any two enemies, there is no person that they are both friends with. In how many ways is this possible? | 52 | Denote friendship between two people $a$ and $b$ by $a \sim b$. Then, assuming everyone is friends with themselves, the following conditions are satisfied: ・ $a \sim a$ - If $a \sim b$, then $b \sim a$ - If $a \sim b$ and $b \sim c$, then $a \sim c$ Thus we can separate the five people into a few groups (possibly one g... | 5.875 | [
5,
6,
6,
6,
6,
6,
6,
6
] |
In a group of people, there are 13 who like apples, 9 who like blueberries, 15 who like cantaloupe, and 6 who like dates. (A person can like more than 1 kind of fruit.) Each person who likes blueberries also likes exactly one of apples and cantaloupe. Each person who likes cantaloupe also likes exactly one of blueberri... | 22 | Everyone who likes cantaloupe likes exactly one of blueberries and dates. However, there are 15 people who like cantaloupe, 9 who like blueberries, and 6 who like dates. Thus, everyone who likes blueberries or dates must also like cantaloupes (because if any of them didn't, we would end up with less than 15 people who ... | 5.5 | [
5,
5,
6,
5,
6,
6,
5,
6
] |
A triangle in the $x y$-plane is such that when projected onto the $x$-axis, $y$-axis, and the line $y=x$, the results are line segments whose endpoints are $(1,0)$ and $(5,0),(0,8)$ and $(0,13)$, and $(5,5)$ and $(7.5,7.5)$, respectively. What is the triangle's area? | \frac{17}{2} | Sketch the lines $x=1, x=5, y=8, y=13, y=10-x$, and $y=15-x$. The triangle has to be contained in the hexagonal region contained in all these lines. If all the projections are correct, every other vertex of the hexagon must be a vertex of the triangle, which gives us two possibilities for the triangle. One of these tri... | 5.625 | [
5,
6,
6,
6,
6,
5,
5,
6
] |
Kelvin the Frog is trying to hop across a river. The river has 10 lilypads on it, and he must hop on them in a specific order (the order is unknown to Kelvin). If Kelvin hops to the wrong lilypad at any point, he will be thrown back to the wrong side of the river and will have to start over. Assuming Kelvin is infinite... | 176 | Kelvin needs (at most) $i(10-i)$ hops to determine the $i$ th lilypad he should jump to, then an additional 11 hops to actually get across the river. Thus he requires $\sum_{i=1}^{10} i(10-i)+11=176$ hops to guarantee success. | 4.75 | [
4,
5,
5,
5,
5,
5,
5,
4
] |
Indecisive Andy starts out at the midpoint of the 1-unit-long segment $\overline{H T}$. He flips 2010 coins. On each flip, if the coin is heads, he moves halfway towards endpoint $H$, and if the coin is tails, he moves halfway towards endpoint $T$. After his 2010 moves, what is the expected distance between Andy and th... | \frac{1}{4} | Let Andy's position be $x$ units from the H end after 2009 flips. If Any moves towards the $H$ end, he ends up at $\frac{x}{2}$, a distance of $\frac{1-x}{2}$ from the midpoint. If Andy moves towards the $T$ end, he ends up at $\frac{1+x}{2}$, a distance of $\frac{x}{2}$ from the midpoint. His expected distance from th... | 5.25 | [
5,
4,
5,
6,
6,
6,
4,
6
] |
In convex quadrilateral \(ABCD\) with \(AB=11\) and \(CD=13\), there is a point \(P\) for which \(\triangle ADP\) and \(\triangle BCP\) are congruent equilateral triangles. Compute the side length of these triangles. | 7 | Evidently \(ABCD\) is an isosceles trapezoid with \(P\) as its circumcenter. Now, construct isosceles trapezoid \(ABB'C\) (that is, \(BB'\) is parallel to \(AC\).) Then \(AB'PD\) is a rhombus, so \(\angle B'CD=\frac{1}{2} \angle B'PD=60^{\circ}\) by the inscribed angle theorem. Also, \(B'C=11\) because the quadrilatera... | 5.375 | [
5,
6,
5,
5,
6,
6,
5,
5
] |
Let $Q(x)=x^{2}+2x+3$, and suppose that $P(x)$ is a polynomial such that $P(Q(x))=x^{6}+6x^{5}+18x^{4}+32x^{3}+35x^{2}+22x+8$. Compute $P(2)$. | 2 | Note that $Q(-1)=2$. Therefore, $P(2)=P(Q(-1))=1-6+18-32+35-22+8=2$. | 4.25 | [
4,
4,
4,
4,
5,
5,
4,
4
] |
Let \(ABCD\) be a trapezoid such that \(AB \parallel CD, \angle BAC=25^{\circ}, \angle ABC=125^{\circ}\), and \(AB+AD=CD\). Compute \(\angle ADC\). | 70^{\circ} | Construct the parallelogram \(ABED\). From the condition \(AB+AD=CD\), we get that \(EC=AD=EB\). Thus, \(\angle ADC=\angle BEC=180^{\circ}-2 \angle BCE=180^{\circ}-2 \cdot 55^{\circ}=70^{\circ}\). | 5.25 | [
5,
5,
5,
5,
5,
5,
6,
6
] |
Let $a, b$ be positive reals with $a>b>\frac{1}{2} a$. Place two squares of side lengths $a, b$ next to each other, such that the larger square has lower left corner at $(0,0)$ and the smaller square has lower left corner at $(a, 0)$. Draw the line passing through $(0, a)$ and $(a+b, 0)$. The region in the two squares ... | \frac{5}{3} | Let $t=\frac{a}{b} \in(1,2)$; we will rewrite the sum $a+b$ as a function of $t$. The area condition easily translates to $\frac{a^{2}-a b+2 b^{2}}{2}=2013$, or $b^{2}\left(t^{2}-t+2\right)=4026 \Longleftrightarrow b=\sqrt{\frac{4026}{t^{2}-t+2}}$. Thus $a+b$ is a function $f(t)=(1+t) \sqrt{\frac{4026}{t^{2}-t+2}}$ of ... | 7.5 | [
7,
7,
8,
8,
7,
8,
7,
8
] |
On an $8 \times 8$ chessboard, 6 black rooks and $k$ white rooks are placed on different cells so that each rook only attacks rooks of the opposite color. Compute the maximum possible value of $k$. | 14 | The answer is $k=14$. For a valid construction, place the black rooks on cells $(a, a)$ for $2 \leq a \leq 7$ and the white rooks on cells $(a, a+1)$ and $(a+1, a)$ for $1 \leq a \leq 7$. Now, we prove the optimality. As rooks can only attack opposite color rooks, the color of rooks in each row is alternating. The diff... | 5.625 | [
6,
6,
5,
6,
6,
5,
5,
6
] |
A polygon \(\mathcal{P}\) is drawn on the 2D coordinate plane. Each side of \(\mathcal{P}\) is either parallel to the \(x\) axis or the \(y\) axis (the vertices of \(\mathcal{P}\) do not have to be lattice points). Given that the interior of \(\mathcal{P}\) includes the interior of the circle \(x^{2}+y^{2}=2022\), find... | 8 \sqrt{2022} | The minimum possible perimeter is achieved by an axis-aligned square with all four sides tangent to the circle, which has area \(8 \sqrt{2022}\). To see why this is true, notice that there must be at least \(2 \sqrt{2022}\) length of total perimeter facing left, \(2 \sqrt{2022}\) length facing up, \(2 \sqrt{2022}\) fac... | 5.5 | [
5,
6,
5,
6,
6,
5,
6,
5
] |
Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament will have 4 rounds, and in each round the 9 players are split into 3 groups of 3. During the tournament, each player plays every other player exactly once. How many different ways can Danielle divide the 9 people into three group... | 20160 | We first split the 9 people up arbitrarily into groups of 3. There are $\frac{\binom{9}{3}\binom{6}{3}\binom{3}{3}}{3!}=280$ ways of doing this. Without loss of generality, label the people 1 through 9 so that the first round groups are $\{1,2,3\},\{4,5,6\}$, and $\{7,8,9\}$. We will use this numbering to count the num... | 6.5 | [
6,
7,
7,
7,
6,
6,
6,
7
] |
Find the volume of the set of points $(x, y, z)$ satisfying $$\begin{array}{r} x, y, z \geq 0 \\ x+y \leq 1 \\ y+z \leq 1 \\ z+x \leq 1 \end{array}$$ | \frac{1}{4} | Without loss of generality, assume that $x \geq y$ - half the volume of the solid is on this side of the plane $x=y$. For each value of $c$ from 0 to $\frac{1}{2}$, the region of the intersection of this half of the solid with the plane $y=c$ is a trapezoid. The trapezoid has height $1-2 c$ and average base $\frac{1}{2... | 5 | [
5,
5,
6,
5,
4,
6,
5,
4
] |
In the Democratic Republic of Irun, 5 people are voting in an election among 5 candidates. If each person votes for a single candidate at random, what is the expected number of candidates that will be voted for? | \frac{2101}{625} | The probability that a chosen candidate will receive no votes at all is $\left(\frac{4}{5}\right)^{5}$, or the probability that every person will vote for someone other than that one candidate. Then the probability that a chosen candidate will receive at least one vote is $1-\left(\frac{4}{5}\right)^{5}=\frac{2101}{312... | 4.25 | [
5,
4,
4,
4,
4,
5,
4,
4
] |
Marisa has a collection of $2^{8}-1=255$ distinct nonempty subsets of $\{1,2,3,4,5,6,7,8\}$. For each step she takes two subsets chosen uniformly at random from the collection, and replaces them with either their union or their intersection, chosen randomly with equal probability. (The collection is allowed to contain ... | \frac{1024}{255} | It suffices to compute the probability of each number appearing in the final subset. For any given integer $n \in[1,8]$, there are $2^{7}=128$ subsets with $n$ and $2^{7}-1=127$ without. When we focus on only this element, each operation is equivalent to taking two random sets and discarding one of them randomly. There... | 6.875 | [
8,
6,
7,
7,
7,
7,
7,
6
] |
The English alphabet, which has 26 letters, is randomly permuted. Let \(p_{1}\) be the probability that \(\mathrm{AB}, \mathrm{CD}\), and \(\mathrm{EF}\) all appear as contiguous substrings. Let \(p_{2}\) be the probability that \(\mathrm{ABC}\) and \(\mathrm{DEF}\) both appear as contiguous substrings. Compute \(\frac... | 23 | There are 23! ways to arrange the alphabet such that AB, CD, and EF all appear as contiguous substrings: treat each of these pairs of letters as a single merged symbol, which leaves 23 symbols to permute. Similarly, there are 22! ways to arrange the alphabet such that ABC and DEF both appear as contiguous substrings. T... | 5.375 | [
5,
6,
5,
6,
5,
6,
5,
5
] |
We say that a positive real number $d$ is good if there exists an infinite sequence $a_{1}, a_{2}, a_{3}, \ldots \in(0, d)$ such that for each $n$, the points $a_{1}, \ldots, a_{n}$ partition the interval $[0, d]$ into segments of length at most $1 / n$ each. Find $\sup \{d \mid d \text { is good }\}$. | \ln 2 | Let $d^{\star}=\sup \{d \mid d$ is good $\}$. We will show that $d^{\star}=\ln (2) \doteq 0.693$. 1. $d^{\star} \leq \ln 2:$ Assume that some $d$ is good and let $a_{1}, a_{2}, \ldots$ be the witness sequence. Fix an integer $n$. By assumption, the prefix $a_{1}, \ldots, a_{n}$ of the sequence splits the interval $[0, ... | 7.875 | [
8,
8,
8,
8,
7,
8,
8,
8
] |
Alice is once again very bored in class. On a whim, she chooses three primes $p, q, r$ independently and uniformly at random from the set of primes of at most 30. She then calculates the roots of $p x^{2}+q x+r$. What is the probability that at least one of her roots is an integer? | \frac{3}{200} | Since all of the coefficients are positive, any root $x$ must be negative. Moreover, by the rational root theorem, in order for $x$ to be an integer we must have either $x=-1$ or $x=-r$. So we must have either $p r^{2}-q r+r=0 \Longleftrightarrow p r=q-1$ or $p-q+r=0$. Neither of these cases are possible if all three p... | 6 | [
6,
5,
6,
6,
6,
7,
6,
6
] |
Pick a random integer between 0 and 4095, inclusive. Write it in base 2 (without any leading zeroes). What is the expected number of consecutive digits that are not the same (that is, the expected number of occurrences of either 01 or 10 in the base 2 representation)? | \frac{20481}{4096} | Note that every number in the range can be written as a 12-digit binary string. For $i=1,2, \ldots 11$, let $R_{i}$ be a random variable which is 1 if the $i$ th and $(i+1)$ st digits differ in a randomly chosen number in the range. By linearity of expectation, $E\left(\sum_{i} R_{i}\right)=\sum E\left(R_{i}\right)$. S... | 6.375 | [
7,
7,
6,
7,
6,
6,
6,
6
] |
Define a sequence of polynomials as follows: let $a_{1}=3 x^{2}-x$, let $a_{2}=3 x^{2}-7 x+3$, and for $n \geq 1$, let $a_{n+2}=\frac{5}{2} a_{n+1}-a_{n}$. As $n$ tends to infinity, what is the limit of the sum of the roots of $a_{n}$ ? | \frac{13}{3} | By using standard methods for solving linear recurrences, we see that this recurrence has a characteristic polynomial of $x^{2}-\frac{5}{2} x+1=\left(x-\frac{1}{2}\right)(x-2)$, hence $a_{n}(x)=c(x) \cdot 2^{n}+d(x) \cdot 2^{-n}$ for some polynomials $c$ and $d$. Plugging in $n=1$ and $n=2$ gives $$2 c(x)+\frac{1}{2} d... | 6.75 | [
7,
7,
6,
7,
7,
7,
7,
6
] |
From the point $(x, y)$, a legal move is a move to $\left(\frac{x}{3}+u, \frac{y}{3}+v\right)$, where $u$ and $v$ are real numbers such that $u^{2}+v^{2} \leq 1$. What is the area of the set of points that can be reached from $(0,0)$ in a finite number of legal moves? | \frac{9 \pi}{4} | We claim that the set of points is the disc with radius $\frac{3}{2}$ centered at the origin, which clearly has area $\frac{9 \pi}{4}$. First, we show that the set is contained in this disc. This is because if we are currently at a distance of $r$ from the origin, then we can't end up at a distance of greater than $\fr... | 6.25 | [
7,
6,
6,
6,
7,
6,
6,
6
] |
Define the sequence $f_{1}, f_{2}, \ldots:[0,1) \rightarrow \mathbb{R}$ of continuously differentiable functions by the following recurrence: $$ f_{1}=1 ; \quad f_{n+1}^{\prime}=f_{n} f_{n+1} \quad \text { on }(0,1), \quad \text { and } \quad f_{n+1}(0)=1 $$ Show that \(\lim _{n \rightarrow \infty} f_{n}(x)\) exists fo... | \frac{1}{1-x} | First of all, the sequence $f_{n}$ is well defined and it holds that $$ f_{n+1}(x)=e^{\int_{0}^{x} f_{n}(t) \mathrm{d} t} \tag{2} $$ The mapping $\Phi: C([0,1)) \rightarrow C([0,1))$ given by $$ \Phi(g)(x)=e^{\int_{0}^{x} g(t) \mathrm{d} t} $$ is monotone, i.e. if $f<g$ on $(0,1)$ then $$ \Phi(f)(x)=e^{\int_{0}^{x} f(t... | 7.875 | [
8,
8,
8,
7,
8,
8,
8,
8
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.