question_id stringlengths 1 6 | correct_count int64 1 4 | incorrect_count int64 4 7 | total_responses int64 8 8 | problem stringlengths 18 2.09k | answer stringlengths 1 84 | solution stringlengths 0 11.1k |
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3140 | 3 | 5 | 8 | A teacher must divide 221 apples evenly among 403 students. What is the minimal number of pieces into which she must cut the apples? (A whole uncut apple counts as one piece.) | 611 | Consider a bipartite graph, with 221 vertices representing the apples and 403 vertices representing the students; each student is connected to each apple that she gets a piece of. The number of pieces then equals the number of edges in the graph. Each student gets a total of $221 / 403=17 / 31$ apple, but each componen... |
3142 | 4 | 4 | 8 | We are given triangle $A B C$, with $A B=9, A C=10$, and $B C=12$, and a point $D$ on $B C . B$ and $C$ are reflected in $A D$ to $B^{\prime}$ and $C^{\prime}$, respectively. Suppose that lines $B C^{\prime}$ and $B^{\prime} C$ never meet (i.e., are parallel and distinct). Find $B D$. | 6 | The lengths of $A B$ and $A C$ are irrelevant. Because the figure is symmetric about $A D$, lines $B C^{\prime}$ and $B^{\prime} C$ meet if and only if they meet at a point on line $A D$. So, if they never meet, they must be parallel to $A D$. Because $A D$ and $B C^{\prime}$ are parallel, triangles $A B D$ and $A D C^... |
3159 | 2 | 6 | 8 | Find the number of ordered triples of nonnegative integers $(a, b, c)$ that satisfy $(ab+1)(bc+1)(ca+1)=84$. | 12 | The solutions are $(0,1,83)$ and $(1,2,3)$ up to permutation. First, we do the case where at least one of $a, b, c$ is 0. WLOG, say $a=0$. Then we have $1+bc=84 \Longrightarrow bc=83$. As 83 is prime, the only solution is $(0,1,83)$ up to permutation. Otherwise, we claim that at least one of $a, b, c$ is equal to 1. Ot... |
3168 | 4 | 4 | 8 | At a recent math contest, Evan was asked to find $2^{2016}(\bmod p)$ for a given prime number $p$ with $100<p<500$. Evan has forgotten what the prime $p$ was, but still remembers how he solved it: - Evan first tried taking 2016 modulo $p-1$, but got a value $e$ larger than 100. - However, Evan noted that $e-\frac{1}{2}... | 211 | Answer is $p=211$. Let $p=2d+1,50<d<250$. The information in the problem boils down to $2016=d+21 \quad(\bmod 2d)$. From this we can at least read off $d \mid 1995$. Now factor $1995=3 \cdot 5 \cdot 7 \cdot 19$. The values of $d$ in this interval are $57,95,105,133$. The prime values of $2d+1$ are then 191 and 211. Of ... |
3176 | 2 | 6 | 8 | Find the smallest possible value of $x+y$ where $x, y \geq 1$ and $x$ and $y$ are integers that satisfy $x^{2}-29y^{2}=1$ | 11621 | Continued fraction convergents to $\sqrt{29}$ are $5, \frac{11}{2}, \frac{16}{3}, \frac{27}{5}, \frac{70}{13}$ and you get $70^{2}-29 \cdot 13^{2}=-1$ so since $(70+13\sqrt{29})^{2}=9801+1820\sqrt{29}$ the answer is $9801+1820=11621$ |
3178 | 1 | 7 | 8 | Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,...,N\}$ , one can still find $2016$ distinct numbers among the remaining elements with sum $N$ . | \boxed{6097392} | Since any $2016$ elements are removed, suppose we remove the integers from $1$ to $2016$ . Then the smallest possible sum of $2016$ of the remaining elements is \[2017+2018+\cdots + 4032 = 1008 \cdot 6049 = 6097392\] so clearly $N\ge 6097392$ . We will show that $N=6097392$ works.
$\vspace{0.2 in}$
$\{1,2\cdots 609739... |
3186 | 2 | 6 | 8 | Let $a, b, c$ be non-negative real numbers such that $ab+bc+ca=3$. Suppose that $a^{3}b+b^{3}c+c^{3}a+2abc(a+b+c)=\frac{9}{2}$. What is the minimum possible value of $ab^{3}+bc^{3}+ca^{3}$? | 18 | Expanding the inequality $\sum_{\text {cyc }} ab(b+c-2a)^{2} \geq 0$ gives $\left(\sum_{\text {cyc }} ab^{3}\right)+4\left(\sum_{\text {cyc }} a^{3}b\right)-4\left(\sum_{\text {cyc }} a^{2}b^{2}\right)-abc(a+b+c) \geq 0$. Using $\left(\sum_{\text {cyc }} a^{3}b\right)+2abc(a+b+c)=\frac{9}{2}$ in the inequality above yi... |
3193 | 1 | 7 | 8 | Let $ABC$ be a triangle with $AB=5, BC=4$ and $AC=3$. Let $\mathcal{P}$ and $\mathcal{Q}$ be squares inside $ABC$ with disjoint interiors such that they both have one side lying on $AB$. Also, the two squares each have an edge lying on a common line perpendicular to $AB$, and $\mathcal{P}$ has one vertex on $AC$ and $\... | \frac{144}{49} | Let the side lengths of $\mathcal{P}$ and $\mathcal{Q}$ be $a$ and $b$, respectively. Label two of the vertices of $\mathcal{P}$ as $D$ and $E$ so that $D$ lies on $AB$ and $E$ lies on $AC$, and so that $DE$ is perpendicular to $AB$. The triangle $ADE$ is similar to $ACB$. So $AD=\frac{3}{4}a$. Using similar arguments,... |
3197 | 1 | 7 | 8 | A cylinder of base radius 1 is cut into two equal parts along a plane passing through the center of the cylinder and tangent to the two base circles. Suppose that each piece's surface area is $m$ times its volume. Find the greatest lower bound for all possible values of $m$ as the height of the cylinder varies. | 3 | Let $h$ be the height of the cylinder. Then the volume of each piece is half the volume of the cylinder, so it is $\frac{1}{2} \pi h$. The base of the piece has area $\pi$, and the ellipse formed by the cut has area $\pi \cdot 1 \cdot \sqrt{1+\frac{h^{2}}{4}}$ because its area is the product of the semiaxes times $\pi$... |
3201 | 4 | 4 | 8 | Two mathematicians, Kelly and Jason, play a cooperative game. The computer selects some secret positive integer $n<60$ (both Kelly and Jason know that $n<60$, but that they don't know what the value of $n$ is). The computer tells Kelly the unit digit of $n$, and it tells Jason the number of divisors of $n$. Then, Kelly... | 10 | The only way in which Kelly can know that $n$ is divisible by at least two different primes is if she is given 0 as the unit digit of $n$, since if she received anything else, then there is some number with that unit digit and not divisible by two primes (i.e., $1,2,3,4,5,16,7,8,9$ ). Then, after Kelly says the first l... |
3205 | 4 | 4 | 8 | For a positive integer $n$, let $\theta(n)$ denote the number of integers $0 \leq x<2010$ such that $x^{2}-n$ is divisible by 2010. Determine the remainder when $\sum_{n=0}^{2009} n \cdot \theta(n)$ is divided by 2010. | 335 | Let us consider the $\operatorname{sum} \sum_{n=0}^{2009} n \cdot \theta(n)(\bmod 2010)$ in a another way. Consider the sum $0^{2}+1^{2}+2^{2}+\cdots+2007^{2}(\bmod 2010)$. For each $0 \leq n<2010$, in the latter sum, the term $n$ appears $\theta(n)$ times, so the sum is congruent to $\sum_{n=0}^{2009} n \cdot \theta(n... |
3236 | 2 | 6 | 8 | We call a positive integer $t$ good if there is a sequence $a_{0}, a_{1}, \ldots$ of positive integers satisfying $a_{0}=15, a_{1}=t$, and $a_{n-1} a_{n+1}=\left(a_{n}-1\right)\left(a_{n}+1\right)$ for all positive integers $n$. Find the sum of all good numbers. | 296 | By the condition of the problem statement, we have $a_{n}^{2}-a_{n-1} a_{n+1}=1=a_{n-1}^{2}-a_{n-2} a_{n}$. This is equivalent to $\frac{a_{n-2}+a_{n}}{a_{n-1}}=\frac{a_{n-1}+a_{n+1}}{a_{n}}$. Let $k=\frac{a_{0}+a_{2}}{a_{1}}$. Then we have $\frac{a_{n-1}+a_{n+1}}{a_{n}}=\frac{a_{n-2}+a_{n}}{a_{n-1}}=\frac{a_{n-3}+a_{n... |
3238 | 1 | 7 | 8 | Let $A B C$ be a triangle and $\omega$ be its circumcircle. The point $M$ is the midpoint of arc $B C$ not containing $A$ on $\omega$ and $D$ is chosen so that $D M$ is tangent to $\omega$ and is on the same side of $A M$ as $C$. It is given that $A M=A C$ and $\angle D M C=38^{\circ}$. Find the measure of angle $\angl... | 33^{\circ} | By inscribed angles, we know that $\angle B A C=38^{\circ} \cdot 2=76^{\circ}$ which means that $\angle C=104^{\circ}-\angle B$. Since $A M=A C$, we have $\angle A C M=\angle A M C=90^{\circ}-\frac{\angle M A C}{2}=71^{\circ}$. Once again by inscribed angles, this means that $\angle B=71^{\circ}$ which gives $\angle C=... |
3240 | 1 | 7 | 8 | How many ways are there to label the faces of a regular octahedron with the integers 18, using each exactly once, so that any two faces that share an edge have numbers that are relatively prime? Physically realizable rotations are considered indistinguishable, but physically unrealizable reflections are considered diff... | 12 | Well, instead of labeling the faces of a regular octahedron, we may label the vertices of a cube. Then, as no two even numbers may be adjacent, the even numbers better form a regular tetrahedron, which can be done in 2 ways (because rotations are indistinguishable but reflections are different). Then 3 must be opposite... |
3247 | 1 | 7 | 8 | Find all integers $m$ such that $m^{2}+6 m+28$ is a perfect square. | 6,-12 | We must have $m^{2}+6 m+28=n^{2}$, where $n$ is an integer. Rewrite this as $(m+3)^{2}+19=$ $n^{2} \Rightarrow n^{2}-(m+3)^{2}=19 \Rightarrow(n-m-3)(n+m+3)=19$. Let $a=n-m-3$ and $b=n+m+3$, so we want $a b=19$. This leaves only 4 cases: - $a=1, b=19$. Solve the system $n-m-3=1$ and $n+m+3=19$ to get $n=10$ and $m=6$, g... |
3252 | 4 | 4 | 8 | Determine all \(\alpha \in \mathbb{R}\) such that for every continuous function \(f:[0,1] \rightarrow \mathbb{R}\), differentiable on \((0,1)\), with \(f(0)=0\) and \(f(1)=1\), there exists some \(\xi \in(0,1)\) such that \(f(\xi)+\alpha=f^{\prime}(\xi)\). | \(\alpha = \frac{1}{e-1}\) | First consider the function \(h(x)=\frac{e^{x}-1}{e-1}\), which has the property that \(h^{\prime}(x)=\frac{e^{x}}{e-1}\). Note that \(h \in V\) and that \(h^{\prime}(x)-h(x)=1 /(e-1)\) is constant. As such, \(\alpha=1 /(e-1)\) is the only possible value that could possibly satisfy the condition from the problem. For \... |
3259 | 1 | 7 | 8 | In the alphametic $W E \times E Y E=S C E N E$, each different letter stands for a different digit, and no word begins with a 0. The $W$ in this problem has the same value as the $W$ in problem 31. Find $S$. | 5 | Problems 31-33 go together. See below. |
3265 | 3 | 5 | 8 | There are 100 houses in a row on a street. A painter comes and paints every house red. Then, another painter comes and paints every third house (starting with house number 3) blue. Another painter comes and paints every fifth house red (even if it is already red), then another painter paints every seventh house blue, a... | 52 | House $n$ ends up red if and only if the largest odd divisor of $n$ is of the form $4 k+1$. We have 25 values of $n=4 k+1 ; 13$ values of $n=2(4 k+1)$ (given by $k=0,1,2, \ldots, 12$ ); 7 values of $n=4(4 k+1)(k=0,1, \ldots, 6) ; 3$ values of $n=8(4 k+1)(k=0,1,2) ; 2$ of the form $n=16(4 k+1$ ) (for $k=0,1)$; 1 of the ... |
3271 | 1 | 7 | 8 | Let $a \geq b \geq c$ be real numbers such that $$\begin{aligned} a^{2} b c+a b^{2} c+a b c^{2}+8 & =a+b+c \\ a^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b+3 a b c & =-4 \\ a^{2} b^{2} c+a b^{2} c^{2}+a^{2} b c^{2} & =2+a b+b c+c a \end{aligned}$$ If $a+b+c>0$, then compute the integer nearest to $a^{5}$. | 1279 | We factor the first and third givens, obtaining the system $$\begin{aligned} a^{2} b c+a b^{2} c+a b c^{2}-a-b-c=(a b c-1)(a+b+c) & =-8 \\ a^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b+3 a b c=(a b+b c+c a)(a+b+c) & =-4 \\ a^{2} b^{2} c+a b^{2} c^{2}+a^{2} b c^{2}-a b-b c-c a=(a b c-1)(a b+b c+c a) & =2 \end{aligned}... |
3284 | 4 | 4 | 8 | In bridge, a standard 52-card deck is dealt in the usual way to 4 players. By convention, each hand is assigned a number of "points" based on the formula $$4 \times(\# \mathrm{~A} \text { 's })+3 \times(\# \mathrm{~K} \text { 's })+2 \times(\# \mathrm{Q} \text { 's })+1 \times(\# \mathrm{~J} \text { 's })$$ Given that ... | \frac{197}{1820} | Obviously, we can ignore the cards lower than J. Simply enumerate the ways to get at least 13 points: AAAA (1), AAAK (16), AAAQ (16), AAAJ (16), AAKK (36), AAKQ (96), AKKK (16). The numbers in parentheses represent the number of ways to choose the suits, given the choices for the values. We see that there are a total o... |
3292 | 3 | 5 | 8 | Given a point $p$ and a line segment $l$, let $d(p, l)$ be the distance between them. Let $A, B$, and $C$ be points in the plane such that $A B=6, B C=8, A C=10$. What is the area of the region in the $(x, y)$-plane formed by the ordered pairs $(x, y)$ such that there exists a point $P$ inside triangle $A B C$ with $d(... | \frac{288}{5} | Place $A B C$ in the coordinate plane so that $A=(0,6), B=(0,0), C=(8,0)$. Consider a point $P=(a, b)$ inside triangle $A B C$. Clearly, $d(P, A B)=a, d(P, B C)=b$. Now, we see that the area of triangle $A B C$ is $\frac{6 \cdot 8}{2}=24$, but may also be computed by summing the areas of triangles $P A B, P B C, P C A$... |
3297 | 3 | 5 | 8 | Let $W$ be the hypercube $\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid 0 \leq x_{1}, x_{2}, x_{3}, x_{4} \leq 1\right\}$. The intersection of $W$ and a hyperplane parallel to $x_{1}+x_{2}+x_{3}+x_{4}=0$ is a non-degenerate 3-dimensional polyhedron. What is the maximum number of faces of this polyhedron? | 8 | The number of faces in the polyhedron is equal to the number of distinct cells (3-dimensional faces) of the hypercube whose interior the hyperplane intersects. However, it is possible to arrange the hyperplane such that it intersects all 8 cells. Namely, $x_{1}+x_{2}+x_{3}+x_{4}=\frac{3}{2}$ intersects all 8 cells beca... |
3307 | 2 | 6 | 8 | Let $A, B, C, D, E, F$ be 6 points on a circle in that order. Let $X$ be the intersection of $AD$ and $BE$, $Y$ is the intersection of $AD$ and $CF$, and $Z$ is the intersection of $CF$ and $BE$. $X$ lies on segments $BZ$ and $AY$ and $Y$ lies on segment $CZ$. Given that $AX=3, BX=2, CY=4, DY=10, EZ=16$, and $FZ=12$, f... | \frac{77}{6} | Let $XY=z, YZ=x$, and $ZX=y$. By Power of a Point, we have that $3(z+10)=2(y+16), 4(x+12)=10(z+3), \text{ and } 12(x+4)=16(y+2)$. Solving this system gives $XY=\frac{11}{3}$ and $YZ=\frac{14}{3}$ and $ZX=\frac{9}{2}$. Therefore, the answer is $XY+YZ+ZX=\frac{77}{6}$. |
3312 | 2 | 6 | 8 | A sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive reals satisfies $a_{n+1}=\sqrt{\frac{1+a_{n}}{2}}$. Determine all $a_{1}$ such that $a_{i}=\frac{\sqrt{6}+\sqrt{2}}{4}$ for some positive integer $i$. | \frac{\sqrt{2}+\sqrt{6}}{2}, \frac{\sqrt{3}}{2}, \frac{1}{2} | Clearly $a_{1}<1$, or else $1 \leq a_{1} \leq a_{2} \leq a_{3} \leq \ldots$ We can therefore write $a_{1}=\cos \theta$ for some $0<\theta<90^{\circ}$. Note that $\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}$, and $\cos 15^{\circ}=$ $\frac{\sqrt{6}+\sqrt{2}}{4}$. Hence, the possibilities for $a_{1}$ are $\cos 15... |
3324 | 1 | 7 | 8 | Define a sequence $a_{i, j}$ of integers such that $a_{1, n}=n^{n}$ for $n \geq 1$ and $a_{i, j}=a_{i-1, j}+a_{i-1, j+1}$ for all $i, j \geq 1$. Find the last (decimal) digit of $a_{128,1}$. | 4 | By applying the recursion multiple times, we find that $a_{1,1}=1, a_{2, n}=n^{n}+(n+1)^{n+1}$, and $a_{3, n}=n^{n}+2(n+1)^{n+1}+(n+2)^{n+2}$. At this point, we can conjecture and prove by induction that $a_{m, n}=\sum_{k=0}^{m-1}\binom{m-1}{k}(n+k)^{n+k}=\sum_{k \geq 0}\binom{m-1}{k}(n+k)^{n+k}$. (The second expressio... |
3331 | 3 | 5 | 8 | Let $a=\sqrt{17}$ and $b=i \sqrt{19}$, where $i=\sqrt{-1}$. Find the maximum possible value of the ratio $|a-z| /|b-z|$ over all complex numbers $z$ of magnitude 1 (i.e. over the unit circle $|z|=1$ ). | \frac{4}{3} | Let $|a-z| /|b-z|=k$. We wish to determine the minimum and maximum value of $k$. Squaring and expansion give: $|a-z|^{2} =|b-z|^{2} \cdot k^{2} |a|^{2}-2 a \cdot z+1 =\left(|b|^{2}-2 b \cdot z+1\right) k^{2} |a|^{2}+1-\left(|b|^{2}+1\right) k^{2} =2\left(a-b k^{2}\right) \cdot z$ where $\cdot$ is a dot product of compl... |
3333 | 1 | 7 | 8 | Let $S=\{1,2,4,8,16,32,64,128,256\}$. A subset $P$ of $S$ is called squarely if it is nonempty and the sum of its elements is a perfect square. A squarely set $Q$ is called super squarely if it is not a proper subset of any squarely set. Find the number of super squarely sets. | 5 | Clearly we may biject squarely sets with binary representations of perfect squares between 1 and $2^{0}+\cdots+2^{8}=2^{9}-1=511$, so there are 22 squarely sets, corresponding to $n^{2}$ for $n=1,2, \ldots, 22$. For convenience, we say $N$ is (super) squarely if and only if the set corresponding to $N$ is (super) squar... |
3336 | 3 | 5 | 8 | Start by writing the integers $1,2,4,6$ on the blackboard. At each step, write the smallest positive integer $n$ that satisfies both of the following properties on the board. - $n$ is larger than any integer on the board currently. - $n$ cannot be written as the sum of 2 distinct integers on the board. Find the 100-th ... | 388 | The sequence goes $1,2,4,6,9,12,17,20,25, \ldots$. Common differences are $5,3,5,3,5,3, \ldots$, starting from 12. Therefore, the answer is $12+47 \times 8=388$. |
3352 | 2 | 6 | 8 | Find the sum of squares of all distinct complex numbers $x$ satisfying the equation $0=4 x^{10}-7 x^{9}+5 x^{8}-8 x^{7}+12 x^{6}-12 x^{5}+12 x^{4}-8 x^{3}+5 x^{2}-7 x+4$ | -\frac{7}{16} | For convenience denote the polynomial by $P(x)$. Notice $4+8=7+5=12$ and that the consecutive terms $12 x^{6}-12 x^{5}+12 x^{4}$ are the leading terms of $12 \Phi_{14}(x)$, which is suggestive. Indeed, consider $\omega$ a primitive 14 -th root of unity; since $\omega^{7}=-1$, we have $4 \omega^{10}=-4 \omega^{3},-7 \om... |
3355 | 3 | 5 | 8 | I have chosen five of the numbers $\{1,2,3,4,5,6,7\}$. If I told you what their product was, that would not be enough information for you to figure out whether their sum was even or odd. What is their product? | 420 | Giving you the product of the five numbers is equivalent to telling you the product of the two numbers I didn't choose. The only possible products that are achieved by more than one pair of numbers are $12(\{3,4\}$ and $\{2,6\})$ and $6(\{1,6\}$ and $\{2,3\})$. But in the second case, you at least know that the two unc... |
3357 | 4 | 4 | 8 | Let $A B C D E$ be a convex pentagon such that $\angle A B C=\angle A C D=\angle A D E=90^{\circ}$ and $A B=B C=C D=D E=1$. Compute $A E$. | 2 | By Pythagoras, $A E^{2}=A D^{2}+1=A C^{2}+2=A B^{2}+3=4$ so $A E=2$. |
3358 | 1 | 7 | 8 | A regular dodecahedron is projected orthogonally onto a plane, and its image is an $n$-sided polygon. What is the smallest possible value of $n$ ? | 6 | We can achieve 6 by projecting onto a plane perpendicular to an edge of the dodecaheron. Indeed, if we imagine viewing the dodecahedron in such a direction, then 4 of the faces are projected to line segments (namely, the two faces adjacent to the edge and the two opposite faces), and of the remaining 8 faces, 4 appear ... |
3361 | 2 | 6 | 8 | We have an $n$-gon, and each of its vertices is labeled with a number from the set $\{1, \ldots, 10\}$. We know that for any pair of distinct numbers from this set there is at least one side of the polygon whose endpoints have these two numbers. Find the smallest possible value of $n$. | 50 | Each number be paired with each of the 9 other numbers, but each vertex can be used in at most 2 different pairs, so each number must occur on at least $\lceil 9 / 2\rceil=5$ different vertices. Thus, we need at least $10 \cdot 5=50$ vertices, so $n \geq 50$. To see that $n=50$ is feasible, let the numbers $1, \ldots, ... |
3362 | 1 | 7 | 8 | $P$ is a point inside triangle $A B C$, and lines $A P, B P, C P$ intersect the opposite sides $B C, C A, A B$ in points $D, E, F$, respectively. It is given that $\angle A P B=90^{\circ}$, and that $A C=B C$ and $A B=B D$. We also know that $B F=1$, and that $B C=999$. Find $A F$. | 499 / 500 | Let $A C=B C=s, A B=B D=t$. Since $B P$ is the altitude in isosceles triangle $A B D$, it bisects angle $B$. So, the Angle Bisector Theorem in triangle $A B C$ given $A E / E C=A B / B C=t / s$. Meanwhile, $C D / D B=(s-t) / t$. Now Ceva's theorem gives us $$ \begin{gathered} \frac{A F}{F B}=\left(\frac{A E}{E C}\right... |
3366 | 3 | 5 | 8 | Mark has a cursed six-sided die that never rolls the same number twice in a row, and all other outcomes are equally likely. Compute the expected number of rolls it takes for Mark to roll every number at least once. | \frac{149}{12} | Suppose Mark has already rolled $n$ unique numbers, where $1 \leq n \leq 5$. On the next roll, there are 5 possible numbers he could get, with $6-n$ of them being new. Therefore, the probability of getting another unique number is $\frac{6-n}{5}$, so the expected number of rolls before getting another unique number is ... |
3370 | 1 | 7 | 8 | Five points are chosen uniformly at random on a segment of length 1. What is the expected distance between the closest pair of points? | \frac{1}{24} | Choose five points arbitrarily at $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ in increasing order. Then the intervals $\left(a_{2}-x, a_{2}\right),\left(a_{3}-x, a_{3}\right),\left(a_{4}-x, a_{4}\right),\left(a_{5}-x, a_{5}\right)$ must all be unoccupied. The probability that this happens is the same as doing the process in re... |
3378 | 1 | 7 | 8 | Let $a \star b=ab-2$. Compute the remainder when $(((579 \star 569) \star 559) \star \cdots \star 19) \star 9$ is divided by 100. | 29 | Note that $$(10a+9) \star (10b+9)=(100ab+90a+90b+81)-2 \equiv 90(a+b)+79 \pmod{100}$$ so throughout our process all numbers will end in 9, so we will just track the tens digit. Then the "new operation" is $$a \dagger b \equiv -(a+b)+7 \bmod 10$$ where $a$ and $b$ track the tens digits. Now $$(a \dagger b) \dagger c \eq... |
3380 | 3 | 5 | 8 | Find the number of ordered pairs of positive integers $(x, y)$ with $x, y \leq 2020$ such that $3 x^{2}+10 x y+3 y^{2}$ is the power of some prime. | 29 | We can factor as $(3 x+y)(x+3 y)$. If $x \geq y$, we need $\frac{3 x+y}{x+3 y} \in\{1,2\}$ to be an integer. So we get the case where $x=y$, in which we need both to be a power of 2, or the case $x=5 y$, in which case we need $y$ to be a power of 2. This gives us $11+9+9=29$ solutions, where we account for $y=5 x$ as w... |
3385 | 4 | 4 | 8 | Compute the number of even positive integers $n \leq 2024$ such that $1,2, \ldots, n$ can be split into $\frac{n}{2}$ pairs, and the sum of the numbers in each pair is a multiple of 3. | 675 | There have to be an even number of multiples of 3 at most $n$, so this means that $n \equiv 0,2 \pmod{6}$. We claim that all these work. We know there are an even number of multiples of 3, so we can pair them; then we can pair $3k+1$ and $3k+2$ for all $k$. This means the answer is $\frac{2022}{3}+1=675$. |
3388 | 1 | 7 | 8 | You would like to provide airline service to the 10 cities in the nation of Schizophrenia, by instituting a certain number of two-way routes between cities. Unfortunately, the government is about to divide Schizophrenia into two warring countries of five cities each, and you don't know which cities will be in each new ... | 30 | Each city $C$ must be directly connected to at least 6 other cities, since otherwise the government could put $C$ in one country and all its connecting cities in the other country, and there would be no way out of $C$. This means that we have 6 routes for each of 10 cities, counted twice (since each route has two endpo... |
3399 | 4 | 4 | 8 | Let $r_{1}, \ldots, r_{n}$ be the distinct real zeroes of the equation $x^{8}-14 x^{4}-8 x^{3}-x^{2}+1=0$. Evaluate $r_{1}^{2}+\cdots+r_{n}^{2}$ | 8 | Observe that $x^{8}-14 x^{4}-8 x^{3}-x^{2}+1 =\left(x^{8}+2 x^{4}+1\right)-\left(16 x^{4}+8 x^{3}+x^{2}\right) =\left(x^{4}+4 x^{2}+x+1\right)\left(x^{4}-4 x^{2}-x+1\right)$. The polynomial $x^{4}+4 x^{2}+x+1=x^{4}+\frac{15}{4} x^{2}+\left(\frac{x}{2}+1\right)^{2}$ has no real roots. On the other hand, let $P(x)=x^{4}-... |
3410 | 3 | 5 | 8 | Find the number of solutions in positive integers $(k ; a_{1}, a_{2}, \ldots, a_{k} ; b_{1}, b_{2}, \ldots, b_{k})$ to the equation $$a_{1}(b_{1})+a_{2}(b_{1}+b_{2})+\cdots+a_{k}(b_{1}+b_{2}+\cdots+b_{k})=7$$ | 15 | Let $k, a_{1}, \ldots, a_{k}, b_{1}, \ldots, b_{k}$ be a solution. Then $b_{1}, b_{1}+b_{2}, \ldots, b_{1}+\cdots+b_{k}$ is just some increasing sequence of positive integers. Considering the $a_{i}$ as multiplicities, the $a_{i}$ 's and $b_{i}$ 's uniquely determine a partition of 7. Likewise, we can determine $a_{i}$... |
3415 | 1 | 7 | 8 | Niffy's favorite number is a positive integer, and Stebbysaurus is trying to guess what it is. Niffy tells her that when expressed in decimal without any leading zeros, her favorite number satisfies the following: - Adding 1 to the number results in an integer divisible by 210 . - The sum of the digits of the number is... | 1010309 | Note that Niffy's favorite number must end in 9, since adding 1 makes it divisible by 10. Also, the sum of the digits of Niffy's favorite number must be even (because it is equal to twice the number of digits) and congruent to 2 modulo 3 (because adding 1 gives a multiple of 3 ). Furthermore, the sum of digits can be a... |
3422 | 2 | 6 | 8 | An ordered pair $(a, b)$ of positive integers is called spicy if $\operatorname{gcd}(a+b, ab+1)=1$. Compute the probability that both $(99, n)$ and $(101, n)$ are spicy when $n$ is chosen from $\{1,2, \ldots, 2024\}$ uniformly at random. | \frac{96}{595} | We claim that $(a, b)$ is spicy if and only if both $\operatorname{gcd}(a+1, b-1)=1$ and $\operatorname{gcd}(a-1, b+1)=1$. To prove the claim, we note that $$\operatorname{gcd}(a+b, ab+1)=\operatorname{gcd}(a+b, b(-b)+1)=\operatorname{gcd}(a+b, b^{2}-1)$$ Hence, we have $$\begin{aligned} \operatorname{gcd}(a+b, ab+1)=1... |
3425 | 2 | 6 | 8 | Let $a, b$ be integers chosen independently and uniformly at random from the set $\{0,1,2, \ldots, 80\}$. Compute the expected value of the remainder when the binomial coefficient $\binom{a}{b}=\frac{a!}{b!(a-b)!}$ is divided by 3. | \frac{1816}{6561} | By Lucas' Theorem we're looking at $\prod_{i=1}^{4}\binom{a_{i}}{b_{i}}$ where the $a_{i}$ and $b_{i}$ are the digits of $a$ and $b$ in base 3. If any $a_{i}<b_{i}$, then the product is zero modulo 3. Otherwise, the potential residues are $\binom{2}{0}=1,\binom{2}{1}=2,\binom{2}{2}=1,\binom{1}{0}=1,\binom{1}{1}=1,\bino... |
3426 | 1 | 7 | 8 | Let $S$ be the set \{1,2, \ldots, 2012\}. A perfectutation is a bijective function $h$ from $S$ to itself such that there exists an $a \in S$ such that $h(a) \neq a$, and that for any pair of integers $a \in S$ and $b \in S$ such that $h(a) \neq a, h(b) \neq b$, there exists a positive integer $k$ such that $h^{k}(a)=b... | 2 | Note that both $f$ and $g$, when written in cycle notation, must contain exactly one cycle that contains more than 1 element. Assume $f$ has $k$ fixed points, and that the other $2012-k$ elements form a cycle, (of which there are (2011 - $k$ )! ways). Then note that if $f$ fixes $a$ then $f(g(a))=g(f(a))=g(a)$ implies ... |
3434 | 4 | 4 | 8 | An omino is a 1-by-1 square or a 1-by-2 horizontal rectangle. An omino tiling of a region of the plane is a way of covering it (and only it) by ominoes. How many omino tilings are there of a 2-by-10 horizontal rectangle? | 7921 | There are exactly as many omino tilings of a 1-by-$n$ rectangle as there are domino tilings of a 2-by-$n$ rectangle. Since the rows don't interact at all, the number of omino tilings of an $m$-by-$n$ rectangle is the number of omino tilings of a 1-by-$n$ rectangle raised to the $m$ th power, $F_{n}^{m}$. The answer is ... |
3435 | 3 | 5 | 8 | Let $a, b$, and $c$ be real numbers such that $a+b+c=100$, $ab+bc+ca=20$, and $(a+b)(a+c)=24$. Compute all possible values of $bc$. | 224, -176 | We first expand the left-hand-side of the third equation to get $(a+b)(a+c)=a^{2}+ac+ab+bc=24$. From this, we subtract the second equation to obtain $a^{2}=4$, so $a=\pm 2$. If $a=2$, plugging into the first equation gives us $b+c=98$ and plugging into the second equation gives us $2(b+c)+bc=20 \Rightarrow 2(98)+bc=20 ... |
3447 | 2 | 6 | 8 | Two $4 \times 4$ squares are randomly placed on an $8 \times 8$ chessboard so that their sides lie along the grid lines of the board. What is the probability that the two squares overlap? | 529/625 | $529 / 625$. Each square has 5 horizontal $\cdot 5$ vertical $=25$ possible positions, so there are 625 possible placements of the squares. If they do not overlap, then either one square lies in the top four rows and the other square lies in the bottom four rows, or one square lies in the left four columns and the othe... |
3448 | 2 | 6 | 8 | Define the sequence $b_{0}, b_{1}, \ldots, b_{59}$ by $$ b_{i}= \begin{cases}1 & \text { if } \mathrm{i} \text { is a multiple of } 3 \\ 0 & \text { otherwise }\end{cases} $$ Let \left\{a_{i}\right\} be a sequence of elements of \{0,1\} such that $$ b_{n} \equiv a_{n-1}+a_{n}+a_{n+1} \quad(\bmod 2) $$ for $0 \leq n \le... | 0, 3, 5, 6 | Try the four possible combinations of values for $a_{0}$ and $a_{1}$. Since we can write $a_{n} \equiv$ $b_{n-1}-a_{n-2}-a_{n-1}$, these two numbers completely determine the solution $\left\{a_{i}\right\}$ beginning with them (if there is one). For $a_{0}=a_{1}=0$, we can check that the sequence beginning $0,0,0,0,1,1$... |
3451 | 2 | 6 | 8 | Let $P(x)$ be the monic polynomial with rational coefficients of minimal degree such that $\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{4}}, \ldots, \frac{1}{\sqrt{1000}}$ are roots of $P$. What is the sum of the coefficients of $P$? | \frac{1}{16000} | For irrational $\frac{1}{\sqrt{r}},-\frac{1}{\sqrt{r}}$ must also be a root of $P$. Therefore $P(x)=\frac{\left(x^{2}-\frac{1}{2}\right)\left(x^{2}-\frac{1}{3}\right) \cdots\left(x^{2}-\frac{1}{1000}\right)}{\left(x+\frac{1}{2}\right)\left(x+\frac{1}{3}\right) \cdots\left(x+\frac{1}{31}\right)}$. We get the sum of the ... |
3453 | 2 | 6 | 8 | Find the least positive integer $N>1$ satisfying the following two properties: There exists a positive integer $a$ such that $N=a(2 a-1)$. The sum $1+2+\cdots+(N-1)$ is divisible by $k$ for every integer $1 \leq k \leq 10$. | 2016 | The second condition implies that 16 divides $a(2 a-1)\left(2 a^{2}-a-1\right)$, which shows that $a \equiv 0$ or 1 modulo 16. The case $a=1$ would contradict the triviality-avoiding condition $N>1$. $a$ cannot be 16, because 7 does not divide $a(2 a-1)\left(2 a^{2}-a-1\right)$. a cannot be 17, because 9 does not divid... |
3456 | 4 | 4 | 8 | Let $N$ be a three-digit integer such that the difference between any two positive integer factors of $N$ is divisible by 3 . Let $d(N)$ denote the number of positive integers which divide $N$. Find the maximum possible value of $N \cdot d(N)$. | 5586 | We first note that all the prime factors of $n$ must be 1 modulo 3 (and thus 1 modulo 6 ). The smallest primes with this property are $7,13,19, \ldots$ Since $7^{4}=2401>1000$, the number can have at most 3 prime factors (including repeats). Since $7 \cdot 13 \cdot 19=1729>1000$, the most factors $N$ can have is 6 . Co... |
3457 | 1 | 7 | 8 | Consider the two hands of an analog clock, each of which moves with constant angular velocity. Certain positions of these hands are possible (e.g. the hour hand halfway between the 5 and 6 and the minute hand exactly at the 6), while others are impossible (e.g. the hour hand exactly at the 5 and the minute hand exactly... | 143 | 143 We can look at the twelve-hour cycle beginning at midnight and ending just before noon, since during this time, the clock goes through each possible position exactly once. The minute hand has twelve times the angular velocity of the hour hand, so if the hour hand has made $t$ revolutions from its initial position $... |
3471 | 2 | 6 | 8 | Compute the decimal expansion of \sqrt{\pi}$. Your score will be \min (23, k)$, where $k$ is the number of consecutive correct digits immediately following the decimal point in your answer. | 1.77245385090551602729816 \ldots | For this problem, it is useful to know the following square root algorithm that allows for digit-by-digit extraction of \sqrt{x}$ and gives one decimal place of \sqrt{x}$ for each two decimal places of $x$. We will illustrate how to extract the second digit after the decimal point of \sqrt{\pi}$, knowing that \pi=3.141... |
3472 | 4 | 4 | 8 | How many real solutions are there to the equation $|||| x|-2|-2|-2|=|||| x|-3|-3|-3|$? | 6 | 6. The graphs of the two sides of the equation can be graphed on the same plot to reveal six intersection points. |
3477 | 1 | 7 | 8 | Cyclic quadrilateral $A B C D$ has side lengths $A B=1, B C=2, C D=3$ and $D A=4$. Points $P$ and $Q$ are the midpoints of $\overline{B C}$ and $\overline{D A}$. Compute $P Q^{2}$. | \frac{116}{35} | Construct $\overline{A C}, \overline{A Q}, \overline{B Q}, \overline{B D}$, and let $R$ denote the intersection of $\overline{A C}$ and $\overline{B D}$. Because $A B C D$ is cyclic, we have that $\triangle A B R \sim \triangle D C R$ and $\triangle A D R \sim \triangle B C R$. Thus, we may write $A R=4 x, B R=2 x, C R... |
3481 | 1 | 7 | 8 | Tessa has a figure created by adding a semicircle of radius 1 on each side of an equilateral triangle with side length 2, with semicircles oriented outwards. She then marks two points on the boundary of the figure. What is the greatest possible distance between the two points? | 3 | Note that both points must be in different semicircles to reach the maximum distance. Let these points be $M$ and $N$, and $O_{1}$ and $O_{2}$ be the centers of the two semicircles where they lie respectively. Then $$M N \leq M O_{1}+O_{1} O_{2}+O_{2} N$$ Note that the the right side will always be equal to 3 ($M O_{1}... |
3488 | 1 | 7 | 8 | Find the smallest positive integer $k$ such that $z^{10}+z^{9}+z^{6}+z^{5}+z^{4}+z+1$ divides $z^{k}-1$. | 84 | Let $Q(z)$ denote the polynomial divisor. We need that the roots of $Q$ are $k$-th roots of unity. With this in mind, we might observe that solutions to $z^{7}=1$ and $z \neq 1$ are roots of $Q$, which leads to its factorization. Alternatively, we note that $$(z-1) Q(z)=z^{11}-z^{9}+z^{7}-z^{4}+z^{2}-1=\left(z^{4}-z^{2... |
3496 | 3 | 5 | 8 | The sequence $\left(z_{n}\right)$ of complex numbers satisfies the following properties: $z_{1}$ and $z_{2}$ are not real. $z_{n+2}=z_{n+1}^{2} z_{n}$ for all integers $n \geq 1$. $\frac{z_{n+3}}{z_{n}^{2}}$ is real for all integers $n \geq 1$. $\left|\frac{z_{3}}{z_{4}}\right|=\left|\frac{z_{4}}{z_{5}}\right|=2$ Find ... | 65536 | All complex numbers can be expressed as $r(\cos \theta+i \sin \theta)=r e^{i \theta}$. Let $z_{n}$ be $r_{n} e^{i \theta_{n}}$. $\frac{z_{n+3}}{z_{n}^{2}}=\frac{z_{n+2}^{2} z_{n+1}}{z_{n}^{2}}=\frac{z_{n+1}^{5} z_{n}^{2}}{z_{n}^{2}}=z_{n+1}^{5}$ is real for all $n \geq 1$, so $\theta_{n}=\frac{\pi k_{n}}{5}$ for all $n... |
3512 | 2 | 6 | 8 | Find the number of pentominoes (5-square polyominoes) that span a 3-by-3 rectangle, where polyominoes that are flips or rotations of each other are considered the same polyomino. | 6 | By enumeration, the answer is 6. |
3521 | 4 | 4 | 8 | Let $AD, BE$, and $CF$ be segments sharing a common midpoint, with $AB < AE$ and $BC < BF$. Suppose that each pair of segments forms a $60^{\circ}$ angle, and that $AD=7, BE=10$, and $CF=18$. Let $K$ denote the sum of the areas of the six triangles $\triangle ABC, \triangle BCD, \triangle CDE, \triangle DEF, \triangle ... | 141 | Let $M$ be the common midpoint, and let $x=7, y=10, z=18$. One can verify that hexagon $ABCDEF$ is convex. We have $[ABC]=[ABM]+[BCM]-[ACM]=\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{x}{2} \cdot \frac{y}{2}+\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{y}{2} \cdot \frac{z}{2}-\frac{1}{2} \cdot \frac{\sqrt{3}}{... |
3522 | 3 | 5 | 8 | Two integers are relatively prime if they don't share any common factors, i.e. if their greatest common divisor is 1. Define $\varphi(n)$ as the number of positive integers that are less than $n$ and relatively prime to $n$. Define $\varphi_{d}(n)$ as the number of positive integers that are less than $d n$ and relativ... | 41 | For fixed $n$, the pattern of integers relatively prime to $n$ repeats every $n$ integers, so $\varphi_{d}(n)=d \varphi(n)$. Therefore the expression in the problem equals $\varphi(n)^{3}$. The cube root of 64000 is $40 . \varphi(p)=p-1$ for any prime $p$. Since 40 is one less than a prime, the least $n$ such that $\va... |
3526 | 4 | 4 | 8 | Bob knows that Alice has 2021 secret positive integers $x_{1}, \ldots, x_{2021}$ that are pairwise relatively prime. Bob would like to figure out Alice's integers. He is allowed to choose a set $S \subseteq\{1,2, \ldots, 2021\}$ and ask her for the product of $x_{i}$ over $i \in S$. Alice must answer each of Bob's quer... | 11 | In general, Bob can find the values of all $n$ integers asking only $\left\lfloor\log _{2} n\right\rfloor+1$ queries. For each of Alice's numbers $x_{i}$, let $Q_{i}$ be the set of queries $S$ such that $i \in S$. Notice that all $Q_{i}$ must be nonempty and distinct. If there exists an empty $Q_{i}$, Bob has asked no ... |
3527 | 3 | 5 | 8 | A positive integer $n$ is picante if $n$ ! ends in the same number of zeroes whether written in base 7 or in base 8 . How many of the numbers $1,2, \ldots, 2004$ are picante? | 4 | The number of zeroes in base 7 is the total number of factors of 7 in $1 \cdot 2 \cdots n$, which is $$ \lfloor n / 7\rfloor+\left\lfloor n / 7^{2}\right\rfloor+\left\lfloor n / 7^{3}\right\rfloor+\cdots $$ The number of zeroes in base 8 is $\lfloor a\rfloor$, where $$ a=\left(\lfloor n / 2\rfloor+\left\lfloor n / 2^{2... |
3539 | 1 | 7 | 8 | 2019 points are chosen independently and uniformly at random on the interval $[0,1]$. Tairitsu picks 1000 of them randomly and colors them black, leaving the remaining ones white. Hikari then computes the sum of the positions of the leftmost white point and the rightmost black point. What is the probability that this s... | \frac{1019}{2019} | Note that each point is chosen uniformly and independently from 0 to 1, so we can apply symmetry. Given any coloring, suppose that we flip all the positions of the black points: then the problem becomes computing the probability that the leftmost white point is to the left of the leftmost black point, which is a necess... |
3547 | 2 | 6 | 8 | The numbers $2^{0}, 2^{1}, \cdots, 2^{15}, 2^{16}=65536$ are written on a blackboard. You repeatedly take two numbers on the blackboard, subtract one from the other, erase them both, and write the result of the subtraction on the blackboard. What is the largest possible number that can remain on the blackboard when the... | 131069 | If we reverse the order of the numbers in the final subtraction we perform, then the final number will be negated. Thus, the possible final numbers come in pairs with opposite signs. Therefore, the largest possible number is the negative of the smallest possible number. To get the smallest possible number, clearly we c... |
3549 | 3 | 5 | 8 | Define a power cycle to be a set $S$ consisting of the nonnegative integer powers of an integer $a$, i.e. $S=\left\{1, a, a^{2}, \ldots\right\}$ for some integer $a$. What is the minimum number of power cycles required such that given any odd integer $n$, there exists some integer $k$ in one of the power cycles such th... | 10 | Partition the odd residues mod 1024 into 10 classes: Class 1: $1(\bmod 4)$. Class $n(2 \leq n \leq 9): 2^{n}-1\left(\bmod 2^{n+1}\right)$. Class 10: $-1(\bmod 1024)$. Let $S_{a}$ be the power cycle generated by $a$. If $a$ is in class 1, all of $S_{a}$ is in class 1. If a is in class $n$ $(2 \leq n \leq 9)$, then $S_{a... |
3550 | 4 | 4 | 8 | Let $Z$ be as in problem 15. Let $X$ be the greatest integer such that $|X Z| \leq 5$. Find $X$. | 2 | Problems 13-15 go together. See below. |
3552 | 2 | 6 | 8 | Let $P$ be a point selected uniformly at random in the cube $[0,1]^{3}$. The plane parallel to $x+y+z=0$ passing through $P$ intersects the cube in a two-dimensional region $\mathcal{R}$. Let $t$ be the expected value of the perimeter of $\mathcal{R}$. If $t^{2}$ can be written as $\frac{a}{b}$, where $a$ and $b$ are r... | 12108 | We can divide the cube into 3 regions based on the value of $x+y+z$ which defines the plane: $x+y+z<1,1 \leq x+y+z \leq 2$, and $x+y+z>2$. The two regions on the ends create tetrahedra, each of which has volume $1 / 6$. The middle region is a triangular antiprism with volume $2 / 3$. If our point $P$ lies in the middle... |
3566 | 3 | 5 | 8 | In a 3 by 3 grid of unit squares, an up-right path is a path from the bottom left corner to the top right corner that travels only up and right in steps of 1 unit. For such a path $p$, let $A_{p}$ denote the number of unit squares under the path $p$. Compute the sum of $A_{p}$ over all up-right paths $p$. | 90 | Each path consists of 3 steps up and 3 steps to the right, so there are $\binom{6}{3}=20$ total paths. Consider the sum of the areas of the regions above all of these paths. By symmetry, this is the same as the answer to the problem. For any path, the sum of the areas of the regions above and below it is $3^{2}=9$, so ... |
3576 | 1 | 7 | 8 | On floor 0 of a weird-looking building, you enter an elevator that only has one button. You press the button twice and end up on floor 1. Thereafter, every time you press the button, you go up by one floor with probability $\frac{X}{Y}$, where $X$ is your current floor, and $Y$ is the total number of times you have pre... | 97 | By induction, we can determine that after $n$ total button presses, your current floor is uniformly distributed from 1 to $n-1$ : the base case $n=2$ is trivial to check, and for the $n+1$ th press, the probability that you are now on floor $i$ is $\frac{1}{n-1}\left(1-\frac{i}{n}\right)+\frac{1}{n-1}\left(\frac{i-1}{n... |
3581 | 1 | 7 | 8 | Define $P=\{\mathrm{S}, \mathrm{T}\}$ and let $\mathcal{P}$ be the set of all proper subsets of $P$. (A proper subset is a subset that is not the set itself.) How many ordered pairs $(\mathcal{S}, \mathcal{T})$ of proper subsets of $\mathcal{P}$ are there such that (a) $\mathcal{S}$ is not a proper subset of $\mathcal{... | 7 | For ease of notation, we let $0=\varnothing, 1=\{\mathrm{S}\}, 2=\{\mathrm{T}\}$. Then both $\mathcal{S}$ and $\mathcal{T}$ are proper subsets of $\{0,1,2\}$. We consider the following cases: Case 1. If $\mathcal{S}=\varnothing$, then $\mathcal{S}$ is a proper subset of any set except the empty set, so we must have $\m... |
3584 | 4 | 4 | 8 | Determine the number of subsets $S$ of $\{1,2,3, \ldots, 10\}$ with the following property: there exist integers $a<b<c$ with $a \in S, b \notin S, c \in S$. | 968 | 968. There are $2^{10}=1024$ subsets of $\{1,2, \ldots, 10\}$ altogether. Any subset without the specified property must be either the empty set or a block of consecutive integers. To specify a block of consecutive integers, we either have just one element (10 choices) or a pair of distinct endpoints $\left(\binom{10}{... |
3586 | 2 | 6 | 8 | In how many ways can the numbers $1,2, \ldots, 2002$ be placed at the vertices of a regular 2002-gon so that no two adjacent numbers differ by more than 2? (Rotations and reflections are considered distinct.) | 4004 | 4004. There are 2002 possible positions for the 1. The two numbers adjacent to the 1 must be 2 and 3; there are two possible ways of placing these. The positions of these numbers uniquely determine the rest: for example, if 3 lies clockwise from 1, then the number lying counterclockwise from 2 must be 4; the number lyi... |
3590 | 2 | 6 | 8 | Caroline starts with the number 1, and every second she flips a fair coin; if it lands heads, she adds 1 to her number, and if it lands tails she multiplies her number by 2. Compute the expected number of seconds it takes for her number to become a multiple of 2021. | 4040 | Consider this as a Markov chain on $\mathbb{Z} / 2021 \mathbb{Z}$. This Markov chain is aperiodic (since 0 can go to 0) and any number can be reached from any other number (by adding 1), so it has a unique stationary distribution $\pi$, which is uniform (since the uniform distribution is stationary). It is a well-known... |
3591 | 3 | 5 | 8 | Triangle $ABC$ has side lengths $AB=19, BC=20$, and $CA=21$. Points $X$ and $Y$ are selected on sides $AB$ and $AC$, respectively, such that $AY=XY$ and $XY$ is tangent to the incircle of $\triangle ABC$. If the length of segment $AX$ can be written as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive inte... | 6710 | Note that the incircle of $\triangle ABC$ is the $A$-excenter of $\triangle AXY$. Let $r$ be the radius of this circle. We can compute the area of $\triangle AXY$ in two ways: $$\begin{aligned} K_{AXY} & =\frac{1}{2} \cdot AX \cdot AY \sin A \\ & =r \cdot(AX+AY-XY) / 2 \\ \Longrightarrow AY & =\frac{r}{\sin A} \end{ali... |
3599 | 2 | 6 | 8 | Define a monic irreducible polynomial with integral coefficients to be a polynomial with leading coefficient 1 that cannot be factored, and the prime factorization of a polynomial with leading coefficient 1 as the factorization into monic irreducible polynomials. How many not necessarily distinct monic irreducible poly... | 5 | $x^{8}+x^{4}+1=\left(x^{8}+2 x^{4}+1\right)-x^{4}=\left(x^{4}+1\right)^{2}-\left(x^{2}\right)^{2}=\left(x^{4}-x^{2}+1\right)\left(x^{4}+x^{2}+1\right)=$ $\left(x^{4}-x^{2}+1\right)\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)$, and $x^{8}+x+1=\left(x^{2}+x+1\right)\left(x^{6}-x^{5}+x^{3}-x^{2}+1\right)$. If an integer p... |
3607 | 1 | 7 | 8 | Compute the number of labelings $f:\{0,1\}^{3} \rightarrow\{0,1, \ldots, 7\}$ of the vertices of the unit cube such that $$\left|f\left(v_{i}\right)-f\left(v_{j}\right)\right| \geq d\left(v_{i}, v_{j}\right)^{2}$$ for all vertices $v_{i}, v_{j}$ of the unit cube, where $d\left(v_{i}, v_{j}\right)$ denotes the Euclidean... | 144 | Let $B=\{0,1\}^{3}$, let $E=\{(x, y, z) \in B: x+y+z$ is even $\}$, and let $O=\{(x, y, z) \in B$ : $x+y+z$ is odd $\}$. As all pairs of vertices within $E$ (and within $O$ ) are $\sqrt{2}$ apart, is easy to see that $\{f(E), f(O)\}=\{\{0,2,4,6\},\{1,3,5,7\}\}$. - There are two ways to choose $f(E)$ and $f(O)$; from no... |
3608 | 4 | 4 | 8 | A semicircle with radius 2021 has diameter $AB$ and center $O$. Points $C$ and $D$ lie on the semicircle such that $\angle AOC < \angle AOD = 90^{\circ}$. A circle of radius $r$ is inscribed in the sector bounded by $OA$ and $OC$ and is tangent to the semicircle at $E$. If $CD=CE$, compute $\lfloor r \rfloor$. | 673 | We are given $$m \angle EOC = m \angle COD$$ and $$m \angle AOC + m \angle COD = 2m \angle EOC + m \angle COD = 90^{\circ}$$ So $m \angle EOC = 30^{\circ}$ and $m \angle AOC = 60^{\circ}$. Letting the radius of the semicircle be $R$, we have $$(R-r) \sin \angle AOC = r \Rightarrow r = \frac{1}{3} R$$ so $$\lfloor r \rf... |
3623 | 4 | 4 | 8 | Three points are chosen inside a unit cube uniformly and independently at random. What is the probability that there exists a cube with side length $\frac{1}{2}$ and edges parallel to those of the unit cube that contains all three points? | \frac{1}{8} | Let the unit cube be placed on a $x y z$-coordinate system, with edges parallel to the $x, y, z$ axes. Suppose the three points are labeled $A, B, C$. If there exists a cube with side length $\frac{1}{2}$ and edges parallel to the edges of the unit cube that contain all three points, then there must exist a segment of ... |
3632 | 1 | 7 | 8 | In the base 10 arithmetic problem $H M M T+G U T S=R O U N D$, each distinct letter represents a different digit, and leading zeroes are not allowed. What is the maximum possible value of $R O U N D$? | 16352 | Clearly $R=1$, and from the hundreds column, $M=0$ or 9. Since $H+G=9+O$ or $10+O$, it is easy to see that $O$ can be at most 7, in which case $H$ and $G$ must be 8 and 9, so $M=0$. But because of the tens column, we must have $S+T \geq 10$, and in fact since $D$ cannot be 0 or $1, S+T \geq 12$, which is impossible giv... |
3633 | 1 | 7 | 8 | For positive integers $n$, let $c_{n}$ be the smallest positive integer for which $n^{c_{n}}-1$ is divisible by 210, if such a positive integer exists, and $c_{n}=0$ otherwise. What is $c_{1}+c_{2}+\cdots+c_{210}$? | 329 | In order for $c_{n} \neq 0$, we must have $\operatorname{gcd}(n, 210)=1$, so we need only consider such $n$. The number $n^{c_{n}}-1$ is divisible by 210 iff it is divisible by each of 2, 3, 5, and 7, and we can consider the order of $n$ modulo each modulus separately; $c_{n}$ will simply be the LCM of these orders. We... |
3636 | 1 | 7 | 8 | If 5 points are placed in the plane at lattice points (i.e. points $(x, y)$ where $x$ and $y$ are both integers) such that no three are collinear, then there are 10 triangles whose vertices are among these points. What is the minimum possible number of these triangles that have area greater than $1 / 2$ ? | 4 | By the pigeonhole principle, the 5 points cannot all be distinct modulo 2, so two of them must have a midpoint that is also a lattice point. This midpoint is not one of the 5 since no 3 are collinear. Pick's theorem states that the area of a polygon whose vertices are lattice points is $B / 2+I-1$ where $B$ is the numb... |
3642 | 1 | 7 | 8 | Five people are at a party. Each pair of them are friends, enemies, or frenemies (which is equivalent to being both friends and enemies). It is known that given any three people $A, B, C$ : - If $A$ and $B$ are friends and $B$ and $C$ are friends, then $A$ and $C$ are friends; - If $A$ and $B$ are enemies and $B$ and $... | 17 | If $A$ and $B$ are frenemies, then regardless of whether another person $C$ is friends or enemies with $A$, $C$ will have to be frenemies with $B$ and vice versa. Therefore, if there is one pair of frenemies then all of them are frenemies with each other, and there is only one possibility. If there are no frenemies, th... |
3645 | 1 | 7 | 8 | The sides of a regular hexagon are trisected, resulting in 18 points, including vertices. These points, starting with a vertex, are numbered clockwise as $A_{1}, A_{2}, \ldots, A_{18}$. The line segment $A_{k} A_{k+4}$ is drawn for $k=1,4,7,10,13,16$, where indices are taken modulo 18. These segments define a region co... | 9/13 | Let us assume all sides are of side length 3. Consider the triangle $A_{1} A_{4} A_{5}$. Let $P$ be the point of intersection of $A_{1} A_{5}$ with $A_{4} A_{8}$. This is a vertex of the inner hexagon. Then $\angle A_{4} A_{1} A_{5}=\angle A_{5} A_{4} P$, by symmetry. It follows that $A_{1} A_{4} A_{5} \sim A_{4} P A_{... |
3646 | 1 | 7 | 8 | Find the number of positive integer solutions to $n^{x}+n^{y}=n^{z}$ with $n^{z}<2001$. | 10 | If $n=1$, the relation can not hold, so assume otherwise. If $x>y$, the left hand side factors as $n^{y}\left(n^{x-y}+1\right)$ so $n^{x-y}+1$ is a power of $n$. But it leaves a remainder of 1 when divided by $n$ and is greater than 1, a contradiction. We reach a similar contradiction if $y>x$. So $y=x$ and $2 n^{x}=n^... |
3649 | 3 | 5 | 8 | In a chess-playing club, some of the players take lessons from other players. It is possible (but not necessary) for two players both to take lessons from each other. It so happens that for any three distinct members of the club, $A, B$, and $C$, exactly one of the following three statements is true: $A$ takes lessons ... | 4 | If $P, Q, R, S$, and $T$ are any five distinct players, then consider all pairs $A, B \in$ $\{P, Q, R, S, T\}$ such that $A$ takes lessons from $B$. Each pair contributes to exactly three triples $(A, B, C)$ (one for each of the choices of $C$ distinct from $A$ and $B$ ); three triples $(C, A, B)$; and three triples $(... |
3657 | 2 | 6 | 8 | Find all prime numbers $p$ such that $y^{2}=x^{3}+4x$ has exactly $p$ solutions in integers modulo $p$. In other words, determine all prime numbers $p$ with the following property: there exist exactly $p$ ordered pairs of integers $(x, y)$ such that $x, y \in\{0,1, \ldots, p-1\}$ and $p \text{ divides } y^{2}-x^{3}-4x$... | p=2 \text{ and } p \equiv 3(\bmod 4) | Clearly $p=2$ works with solutions $(0,0)$ and $(1,1)$ and not $(0,1)$ or $(1,0)$. If $p \equiv 3(\bmod 4)$ then -1 is not a quadratic residue, so for $x^{3}+4x \neq 0$, exactly one of $x^{3}+4x$ and $-x^{3}-4x$ is a square and gives two solutions (for positive and negative $y$), so there's exactly two solutions for ea... |
3658 | 4 | 4 | 8 | Suppose there exists a convex $n$-gon such that each of its angle measures, in degrees, is an odd prime number. Compute the difference between the largest and smallest possible values of $n$. | 356 | We can't have $n=3$ since the sum of the angles must be $180^{\circ}$ but the sum of three odd numbers is odd. On the other hand, for $n=4$ we can take a quadrilateral with angle measures $83^{\circ}, 83^{\circ}, 97^{\circ}, 97^{\circ}$. The largest possible value of $n$ is 360. For larger $n$ we can't even have all an... |
3663 | 1 | 7 | 8 | Compute $$2 \sqrt{2 \sqrt[3]{2 \sqrt[4]{2 \sqrt[5]{2 \cdots}}}}$$ | 2^{e-1} | Taking the base 2 logarithm of the expression gives $$1+\frac{1}{2}\left(1+\frac{1}{3}\left(1+\frac{1}{4}(1+\cdots)\right)\right)=1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots=e-1$$ Therefore the expression is just $2^{e-1}$. |
3670 | 1 | 7 | 8 | How many different graphs with 9 vertices exist where each vertex is connected to 2 others? | 4 | It suffices to consider the complements of the graphs, so we are looking for graphs with 9 vertices, where each vertex is connected to 2 others. There are $\mathbf{4}$ different graphs. |
3672 | 4 | 4 | 8 | On the Cartesian plane $\mathbb{R}^{2}$, a circle is said to be nice if its center is at the origin $(0,0)$ and it passes through at least one lattice point (i.e. a point with integer coordinates). Define the points $A=(20,15)$ and $B=(20,16)$. How many nice circles intersect the open segment $A B$ ? | 10 | The square of the radius of a nice circle is the sum of the square of two integers. The nice circle of radius $r$ intersects (the open segment) $\overline{A B}$ if and only if a point on $\overline{A B}$ is a distance $r$ from the origin. $\overline{A B}$ consists of the points $(20, t)$ where $t$ ranges over $(15,16)$... |
3675 | 1 | 7 | 8 | Consider the equation $F O R T Y+T E N+T E N=S I X T Y$, where each of the ten letters represents a distinct digit from 0 to 9. Find all possible values of $S I X T Y$. | 31486 | Since $Y+N+N$ ends in $Y$, $N$ must be 0 or 5. But if $N=5$ then $T+E+E+1$ ends in T, which is impossible, so $N=0$ and $E=5$. Since $F \neq S$ we must have $O=9, R+T+T+1>10$, and $S=F+1$. Now $I \neq 0$, so it must be that $I=1$ and $R+T+T+1>20$. Thus $R$ and $T$ are 6 and 7, 6 and 8, or 7 and 8 in some order. But $X$... |
3677 | 1 | 7 | 8 | For positive integers $a$ and $b$ such that $a$ is coprime to $b$, define $\operatorname{ord}_{b}(a)$ as the least positive integer $k$ such that $b \mid a^{k}-1$, and define $\varphi(a)$ to be the number of positive integers less than or equal to $a$ which are coprime to $a$. Find the least positive integer $n$ such t... | 240 | The maximum order of an element modulo $n$ is the Carmichael function, denoted $\lambda(n)$. The following properties of the Carmichael function are established: - For primes $p>2$ and positive integers $k, \lambda\left(p^{k}\right)=(p-1) p^{k-1}$. - For a positive integer $k$, $$\lambda\left(2^{k}\right)= \begin{cases... |
3680 | 4 | 4 | 8 | Two fair octahedral dice, each with the numbers 1 through 8 on their faces, are rolled. Let $N$ be the remainder when the product of the numbers showing on the two dice is divided by 8. Find the expected value of $N$. | \frac{11}{4} | If the first die is odd, which has $\frac{1}{2}$ probability, then $N$ can be any of $0,1,2,3,4,5,6,7$ with equal probability, because multiplying each element of $\{0, \ldots, 7\}$ with an odd number and taking modulo 8 results in the same numbers, as all odd numbers are relatively prime to 8. The expected value in th... |
3684 | 3 | 5 | 8 | What is the 18 th digit after the decimal point of $\frac{10000}{9899}$ ? | 5 | $\frac{10000}{9899}$ satisfies $100(x-1)=1.01 x$, so each pair of adjacent digits is generated by adding the previous two pairs of digits. So the decimal is $1.01020305081321345590 \ldots$, and the 18 th digit is 5. |
3686 | 4 | 4 | 8 | $A B C$ is an acute triangle with incircle $\omega$. $\omega$ is tangent to sides $\overline{B C}, \overline{C A}$, and $\overline{A B}$ at $D, E$, and $F$ respectively. $P$ is a point on the altitude from $A$ such that $\Gamma$, the circle with diameter $\overline{A P}$, is tangent to $\omega$. $\Gamma$ intersects $\o... | \frac{675}{4} | By the Law of Sines we have $\sin \angle A=\frac{X Y}{A P}=\frac{4}{5}$. Let $I, T$, and $Q$ denote the center of $\omega$, the point of tangency between $\omega$ and $\Gamma$, and the center of $\Gamma$ respectively. Since we are told $A B C$ is acute, we can compute $\tan \angle \frac{A}{2}=\frac{1}{2}$. Since $\angl... |
3687 | 3 | 5 | 8 | Three fair six-sided dice, each numbered 1 through 6 , are rolled. What is the probability that the three numbers that come up can form the sides of a triangle? | 37/72 | Denote this probability by $p$, and let the three numbers that come up be $x, y$, and $z$. We will calculate $1-p$ instead: $1-p$ is the probability that $x \geq y+z, y \geq z+x$, or $z \geq x+y$. Since these three events are mutually exclusive, $1-p$ is just 3 times the probability that $x \geq y+z$. This happens with... |
3693 | 1 | 7 | 8 | Let $A B C$ be a triangle with $A B=13, B C=14, C A=15$. Let $O$ be the circumcenter of $A B C$. Find the distance between the circumcenters of triangles $A O B$ and $A O C$. | \frac{91}{6} | Let $S, T$ be the intersections of the tangents to the circumcircle of $A B C$ at $A, C$ and at $A, B$ respectively. Note that $A S C O$ is cyclic with diameter $S O$, so the circumcenter of $A O C$ is the midpoint of $O S$, and similarly for the other side. So the length we want is $\frac{1}{2} S T$. The circumradius ... |
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