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3695 | 1 | 7 | 8 | Let $m, n > 2$ be integers. One of the angles of a regular $n$-gon is dissected into $m$ angles of equal size by $(m-1)$ rays. If each of these rays intersects the polygon again at one of its vertices, we say $n$ is $m$-cut. Compute the smallest positive integer $n$ that is both 3-cut and 4-cut. | 14 | For the sake of simplicity, inscribe the regular polygon in a circle. Note that each interior angle of the regular $n$-gon will subtend $n-2$ of the $n$ arcs on the circle. Thus, if we dissect an interior angle into $m$ equal angles, then each must be represented by a total of $\frac{n-2}{m}$ arcs. However, since each of the rays also passes through another vertex of the polygon, that means $\frac{n-2}{m}$ is an integer and thus our desired criteria is that $m$ divides $n-2$. That means we want the smallest integer $n>2$ such that $n-2$ is divisible by 3 and 4 which is just $12+2=14$. |
3707 | 1 | 7 | 8 | Let $A=\{a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{10}, b_{10}\}$, and consider the 2-configuration $C$ consisting of \( \{a_{i}, b_{i}\} \) for all \( 1 \leq i \leq 10, \{a_{i}, a_{i+1}\} \) for all \( 1 \leq i \leq 9 \), and \( \{b_{i}, b_{i+1}\} \) for all \( 1 \leq i \leq 9 \). Find the number of subsets of $C$ that are consistent of order 1. | 89 | Let \( A_{n}=\{a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{n}, b_{n}\} \) for \( n \geq 1 \), and consider the 2-configuration \( C_{n} \) consisting of \( \{a_{i}, b_{i}\} \) for all \( 1 \leq i \leq n, \{a_{i}, a_{i+1}\} \) for all \( 1 \leq i \leq n-1 \), and \( \{b_{i}, b_{i+1}\} \) for all \( 1 \leq i \leq n-1 \). Let \( N_{n} \) be the number of subsets of \( C_{n} \) that are consistent of order 1 (call these "matchings" of \( C_{n} \) ). Consider any matching of \( C_{n+2} \). Either \( a_{n+2} \) is paired with \( b_{n+2} \), in which case the remaining elements of our matching form a matching of \( C_{n+1} \); or \( a_{n+2} \) is paired with \( a_{n+1} \), in which case \( b_{n+2} \) must be paired with \( b_{n+1} \), and the remaining elements form a matching of \( C_{n} \). It follows that \( N_{n+2}=N_{n+1}+N_{n} \). By direct calculation, \( N_{1}=1 \) and \( N_{2}=2 \), and now computing successive values of \( N_{n} \) using the recurrence yields \( N_{10}=89 \). |
3721 | 4 | 4 | 8 | Every second, Andrea writes down a random digit uniformly chosen from the set $\{1,2,3,4\}$. She stops when the last two numbers she has written sum to a prime number. What is the probability that the last number she writes down is 1? | 15/44 | Let $p_{n}$ be the probability that the last number she writes down is 1 when the first number she writes down is $n$. Suppose she starts by writing 2 or 4 . Then she can continue writing either 2 or 4 , but the first time she writes 1 or 3 , she stops. Therefore $p_{2}=p_{4}=\frac{1}{2}$. Suppose she starts by writing 1 . Then she stops if she writes 1, 2 , or 4 , but continues if she writes 3 . Therefore $p_{1}=\frac{1}{4}\left(1+p_{3}\right)$. If she starts by writing 3 , then she stops if she writes 2 or 4 and otherwise continues. Therefore $p_{3}=\frac{1}{4}\left(p_{1}+p_{3}\right)=\frac{1}{16}\left(1+5 p_{3}\right)$. Solving gives $p_{3}=\frac{1}{11}$ and $p_{1}=\frac{3}{11}$. The probability we want to find is therefore $\frac{1}{4}\left(p_{1}+p_{2}+p_{3}+p_{4}\right)=\frac{15}{44}$. |
3734 | 1 | 7 | 8 | In an election for the Peer Pressure High School student council president, there are 2019 voters and two candidates Alice and Celia (who are voters themselves). At the beginning, Alice and Celia both vote for themselves, and Alice's boyfriend Bob votes for Alice as well. Then one by one, each of the remaining 2016 voters votes for a candidate randomly, with probabilities proportional to the current number of the respective candidate's votes. For example, the first undecided voter David has a $\frac{2}{3}$ probability of voting for Alice and a $\frac{1}{3}$ probability of voting for Celia. What is the probability that Alice wins the election (by having more votes than Celia)? | \frac{1513}{2017} | Let $P_{n}(m)$ be the probability that after $n$ voters have voted, Alice gets $m$ votes. We show by induction that for $n \geq 3$, the ratio $P_{n}(2): P_{n}(3): \cdots: P_{n}(n-1)$ is equal to $1: 2: \cdots:(n-2)$. We take a base case of $n=3$, for which the claim is obvious. Then suppose the claim holds for $n=k$. Then $P_{k}(m)=\frac{2 m-2}{(k-1)(k-2)}$. Then $$P_{k+1}(i)=\frac{k-i}{k} P_{k}(i)+\frac{i-1}{k} P_{k}(i-1)=\frac{(k-i)(2 i-2)+(i-1)(2 i-4)}{k(k-1)(k-2)}=\frac{2 i-2}{k(k-1)}$$ Also, we can check $P_{k+1}(2)=\frac{2}{k(k-1)}$ and $P_{k+1}(k)=\frac{2}{k}$, so indeed the claim holds for $n=k+1$, and thus by induction our claim holds for all $n \geq 3$. The probability that Ceila wins the election is then $$\frac{\sum_{m=2}^{1009} P_{2019}(m)}{\sum_{m=2}^{2018} P_{2019}(m)}=\frac{1008 \cdot(1+1008) / 2}{2017 \cdot(1+2017) / 2}=\frac{504}{2017}$$ and thus the probability that Alice wins is $\frac{1513}{2017}$. |
3737 | 1 | 7 | 8 | Regular tetrahedron $A B C D$ is projected onto a plane sending $A, B, C$, and $D$ to $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$ respectively. Suppose $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a convex quadrilateral with $A^{\prime} B^{\prime}=A^{\prime} D^{\prime}$ and $C^{\prime} B^{\prime}=C^{\prime} D^{\prime}$, and suppose that the area of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}=4$. Given these conditions, the set of possible lengths of $A B$ consists of all real numbers in the interval $[a, b)$. Compute $b$. | 2 \sqrt[4]{6} | The value of $b$ occurs when the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ degenerates to an isosceles triangle. This occurs when the altitude from $A$ to $B C D$ is parallel to the plane. Let $s=A B$. Then the altitude from $A$ intersects the center $E$ of face $B C D$. Since $E B=\frac{s}{\sqrt{3}}$, it follows that $A^{\prime} C^{\prime}=A E=\sqrt{s^{2}-\frac{s^{2}}{3}}=\frac{s \sqrt{6}}{3}$. Then since $B D$ is parallel to the plane, $B^{\prime} D^{\prime}=s$. Then the area of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is $4=\frac{1}{2} \cdot \frac{s^{2} \sqrt{6}}{3}$, implying $s^{2}=4 \sqrt{6}$, or $s=2 \sqrt[4]{6}$. |
3750 | 4 | 4 | 8 | A permutation of \{1,2, \ldots, 7\} is chosen uniformly at random. A partition of the permutation into contiguous blocks is correct if, when each block is sorted independently, the entire permutation becomes sorted. For example, the permutation $(3,4,2,1,6,5,7)$ can be partitioned correctly into the blocks $[3,4,2,1]$ and $[6,5,7]$, since when these blocks are sorted, the permutation becomes $(1,2,3,4,5,6,7)$. Find the expected value of the maximum number of blocks into which the permutation can be partitioned correctly. | \frac{151}{105} | Let $\sigma$ be a permutation on \{1, \ldots, n\}. Call $m \in\{1, \ldots, n\}$ a breakpoint of $\sigma$ if $\{\sigma(1), \ldots, \sigma(m)\}=$ $\{1, \ldots, m\}$. Notice that the maximum partition is into $k$ blocks, where $k$ is the number of breakpoints: if our breakpoints are $m_{1}, \ldots, m_{k}$, then we take $\left\{1, \ldots, m_{1}\right\},\left\{m_{1}+1, \ldots, m_{2}\right\}, \ldots,\left\{m_{k-1}+1, \ldots, m_{k}\right\}$ as our contiguous blocks. Now we just want to find $$\mathbb{E}[k]=\mathbb{E}\left[X_{1}+\cdots+X_{n}\right]$$ where $X_{i}=1$ if $i$ is a breakpoint, and $X_{i}=0$ otherwise. We use linearity of expectation and notice that $$\mathbb{E}\left[X_{i}\right]=\frac{i!(n-i)!}{n!}$$ since this is the probability that the first $i$ numbers are just $1, \ldots, i$ in some order. Thus, $$\mathbb{E}[k]=\sum_{i=1}^{n} \frac{i!(n-i)!}{n!}=\sum_{i=1}^{n}\binom{n}{i}^{-1}$$ We can compute for $n=7$ that the answer is $\frac{151}{105}$. |
3752 | 1 | 7 | 8 | For each positive real number $\alpha$, define $$ \lfloor\alpha \mathbb{N}\rfloor:=\{\lfloor\alpha m\rfloor \mid m \in \mathbb{N}\} $$ Let $n$ be a positive integer. A set $S \subseteq\{1,2, \ldots, n\}$ has the property that: for each real $\beta>0$, $$ \text { if } S \subseteq\lfloor\beta \mathbb{N}\rfloor \text {, then }\{1,2, \ldots, n\} \subseteq\lfloor\beta \mathbb{N}\rfloor $$ Determine, with proof, the smallest possible size of $S$. | \lfloor n / 2\rfloor+1 | Answer: $\lfloor n / 2\rfloor+1$ Solution: For each $k \in\{\lceil n / 2\rceil, \ldots, n\}$, picking $\beta=1+1 / k$ gives $$ \lfloor\beta \mathbb{N}\rfloor \cap[n]=[n] \backslash\{k\} $$ so $S$ must contain $k$. Now we show that $S=\{\lceil n / 2\rceil, \ldots, n\}$ works; this set $S$ has $\lfloor n / 2\rfloor+1$ elements. Suppose $\beta$ satisfy $S \subseteq\lfloor\beta \mathbb{N}\rfloor$, and suppose for the sake of contradiction that $[n] \not \subset\lfloor\beta \mathbb{N}\rfloor$. Since we may increase $\beta$ by a small amount $\varepsilon$ without affecting $\lfloor\beta \mathbb{N}\rfloor \cap[n]$, we may assume $\beta$ is irrational. Let $\alpha$ satisfy $1 / \alpha+1 / \beta=1$. By Beatty's Theorem, $\lfloor\alpha \mathbb{N}\rfloor$ and $\lfloor\beta \mathbb{N}\rfloor$ are complement sets in $\mathbb{N}$. Let $m$ be the maximal element of $[n]$ that is not in $\lfloor\beta \mathbb{N}\rfloor$. Then $m=\lfloor k \alpha\rfloor$ for some integer $k$. Consider $m^{\prime}=\lfloor 2 k \alpha\rfloor \in\{2 m, 2 m+1\}$, which must be an element of $\lfloor\alpha \mathbb{N}\rfloor$. Clearly, $m^{\prime}>m$, and since $m<n / 2$, $m^{\prime} \leqslant n$, so $m^{\prime}$ is also an element of $[n]$ that is not in $\lfloor\beta \mathbb{N}\rfloor$. This contradicts the maximality of $m$, and we are done. |
3753 | 1 | 7 | 8 | Determine the number of triples $0 \leq k, m, n \leq 100$ of integers such that $$ 2^{m} n-2^{n} m=2^{k} $$ | 22 | First consider when $n \geq m$, so let $n=m+d$ where $d \geq 0$. Then we have $2^{m}\left(m+d-2^{d} m\right)=$ $2^{m}\left(m\left(1-2^{d}\right)+d\right)$, which is non-positive unless $m=0$. So our first set of solutions is $m=0, n=2^{j}$. Now, we can assume that $m>n$, so let $m=n+d$ where $d>0$. Rewrite $2^{m} n-2^{n} m=2^{n+d} n-$ $2^{n}(n+d)=2^{n}\left(\left(2^{d}-1\right) n-d\right)$. In order for this to be a power of $2,\left(2^{d}-1\right) n-d$ must be a power of 2. This implies that for some $j, 2^{j} \equiv-d\left(\bmod 2^{d}-1\right)$. But notice that the powers of $2\left(\bmod 2^{d}-1\right)$ are $1,2,4, \ldots, 2^{d-1}\left(2^{d} \equiv 1\right.$ so the cycle repeats $)$. In order for the residues to match, we need $2^{j}+d=c\left(2^{d}-1\right)$, where $0 \leq j \leq d-1$ and $c \geq 1$. In order for this to be true, we must have $2^{d-1}+d \geq 2^{d}-1 \Longleftrightarrow d+1 \geq 2^{d-1}$. This inequality is only true for $d=1,2,3$. We plug each of these into the original expression $\left(2^{d}-1\right) n-d$. For $d=1: n-1$ is a power of 2 . This yields the set of solutions $\left(2^{j}+2,2^{j}+1\right)$ for $j \geq 0$. For $d=2: 3 n-2$ is a power of 2 . Note that powers of 2 are $-2(\bmod 3)$ if and only if it is an even power, so $n=\frac{2^{2 j}+2}{3}$. This yields the solution set $\left(\frac{2^{2 j}+8}{3}, \frac{2^{2 j}+2}{3}\right), j \geq 0$. For $d=3: 7 n-3$ is a power of 2 . Powers of 2 have a period of 3 when taken $(\bmod 7)$, so inspection tells us $7 n-3=2^{3 j+2}$, yielding the solution set $\left(\frac{2^{3 j+2}+24}{7}, \frac{2^{3 j+2}+3}{7}\right), j \geq 0$. Therefore, all the solutions are of the form $$ \begin{gathered} (m, n)=\left(0,2^{j}\right),\left(2^{j}+2,2^{j}+1\right) \\ \left(\frac{2^{2 j}+8}{3}, \frac{2^{2 j}+2}{3}\right),\left(\frac{2^{3 j+2}+24}{7}, \frac{2^{3 j+2}+3}{7}\right) \end{gathered} $$ for $j \geq 0$. Restricting this family to $m, n \leq 100$ gives $7+7+5+3=22$. |
3758 | 3 | 5 | 8 | Alice and Bob play a game on a circle with 8 marked points. Alice places an apple beneath one of the points, then picks five of the other seven points and reveals that none of them are hiding the apple. Bob then drops a bomb on any of the points, and destroys the apple if he drops the bomb either on the point containing the apple or on an adjacent point. Bob wins if he destroys the apple, and Alice wins if he fails. If both players play optimally, what is the probability that Bob destroys the apple? | \frac{1}{2} | Let the points be $0, \ldots, 7(\bmod 8)$, and view Alice's reveal as revealing the three possible locations of the apple. If Alice always picks $0,2,4$ and puts the apple randomly at 0 or 4 , by symmetry Bob cannot achieve more than $\frac{1}{2}$. Here's a proof that $\frac{1}{2}$ is always possible. Among the three revealed indices $a, b, c$, positioned on a circle, two must (in the direction in which they're adjacent) have distance at least 3 , so without loss of generality the three are $0, b, c$ where $1 \leq b<c \leq 5$. Modulo reflection and rotation, the cases are: $(0,1,2)$ : Bob places at 1 and wins. $(0,1,3)$ : Bob places at 1 half the time and 3 half the time, so wherever the apple is Bob wins with probability $\frac{1}{2}$. $(0,1,4)$ : Bob places at 1 or 4 , same as above. $(0,2,4)$ : Bob places at 1 or 3 , same as above. $(0,2,5)$ : Bob places at 1 or 5 , same as above. These cover all cases, so we're done. |
3766 | 3 | 5 | 8 | We have a calculator with two buttons that displays an integer $x$. Pressing the first button replaces $x$ by $\left\lfloor\frac{x}{2}\right\rfloor$, and pressing the second button replaces $x$ by $4 x+1$. Initially, the calculator displays 0. How many integers less than or equal to 2014 can be achieved through a sequence of arbitrary button presses? (It is permitted for the number displayed to exceed 2014 during the sequence. Here, $\lfloor y\rfloor$ denotes the greatest integer less than or equal to the real number $y$.) | 233 | We consider the integers from this process written in binary. The first operation truncates the rightmost digit, while the second operation appends 01 to the right. We cannot have a number with a substring 11. For simplicity, call a string valid if it has no consecutive $1^{\prime} s$. Note that any number generated by this process is valid, as truncating the rightmost digit and appending 01 to the right of the digits clearly preserve validity. Since we can effectively append a zero by applying the second operation and then the first operation, we see that we can achieve all valid strings. Note that 2014 has eleven digits when written in binary, and any valid binary string with eleven digits is at most $10111111111=1535$. Therefore, our problem reduces to finding the number of eleven-digit valid strings. Let $F_{n}$ denote the number of valid strings of length $n$. For any valid string of length $n$, we can create a valid string of length $n+1$ by appending a 0, or we can create a valid string of length $n+2$ by appending 01. This process is clearly reversible, so our recursion is given by $F_{n}=F_{n-1}+F_{n-2}$, with $F_{1}=2, F_{2}=3$. This yields a sequence of Fibonacci numbers starting from 2, and some computation shows that our answer is $F_{11}=233$. |
3781 | 3 | 5 | 8 | Calvin has a bag containing 50 red balls, 50 blue balls, and 30 yellow balls. Given that after pulling out 65 balls at random (without replacement), he has pulled out 5 more red balls than blue balls, what is the probability that the next ball he pulls out is red? | \frac{9}{26} | Solution 1. The only information this gives us about the number of yellow balls left is that it is even. A bijection shows that the probability that there are $k$ yellow balls left is equal to the probability that there are $30-k$ yellow balls left (flip the colors of the red and blue balls, and then switch the 65 balls that have been picked with the 65 balls that have not been picked). So the expected number of yellow balls left is 15. Therefore the expected number of red balls left is 22.5. So the answer is $\frac{22.5}{65}=\frac{45}{130}=\frac{9}{26}$. Solution 2. Let $w(b)=\binom{50}{b}\binom{50}{r=b+5}\binom{30}{60-2 b}$ be the number of possibilities in which $b$ blue balls have been drawn (precisely $15 \leq b \leq 30$ are possible). For fixed $b$, the probability of drawing red next is $\frac{50-r}{50+50+30-65}=\frac{45-b}{65}$. So we want to evaluate $$\frac{\sum_{b=15}^{30} w(b) \frac{45-b}{65}}{\sum_{b=15}^{30} w(b)}$$ Note the symmetry of weights: $$w(45-b)=\binom{50}{45-b}\binom{50}{50-b}\binom{30}{2 b-30}=\binom{50}{b+5}\binom{50}{b}\binom{30}{60-2 b}$$ so the $\frac{45-b}{65}$ averages out with $\frac{45-(45-b)}{65}$ to give a final answer of $\frac{45 / 2}{65}=\frac{9}{26}$. Remark. If one looks closely enough, the two approaches are not so different. The second solution may be more conceptually/symmetrically phrased in terms of the number of yellow balls. |
3783 | 4 | 4 | 8 | Compute the number of ways there are to assemble 2 red unit cubes and 25 white unit cubes into a $3 \times 3 \times 3$ cube such that red is visible on exactly 4 faces of the larger cube. (Rotations and reflections are considered distinct.) | 114 | We do casework on the two red unit cubes; they can either be in a corner, an edge, or the center of the face. - If they are both in a corner, they must be adjacent - for each configuration, this corresponds to an edge, of which there are 12. - If one is in the corner and the other is at an edge, we have 8 choices to place the corner. For the edge, the red edge square has to go on the boundary of the faces touching the red corner square, and there are six places here. Thus, we get $8 \cdot 6=48$ configurations. - If one is a corner and the other is in the center of a face, we again have 8 choices for the corner and 3 choices for the center face (the faces not touching the red corner). This gives $8 \cdot 3=24$ options. - We have now completed the cases with a red corner square! Now suppose we have two edges: If we chose in order, we have 12 choices for the first cube. For the second cube, we must place the edge so it covers two new faces, and thus we have five choices. Since we could have picked these edges in either order, we divide by two to avoid overcounting, and we have $12 \cdot 5 / 2=30$ in this case. Now, since edges and faces only cover at most 2 and 1 face respectively, no other configuration works. Thus we have all the cases, and we add: $12+48+24+30=114$. |
3785 | 1 | 7 | 8 | The country of HMMTLand has 8 cities. Its government decides to construct several two-way roads between pairs of distinct cities. After they finish construction, it turns out that each city can reach exactly 3 other cities via a single road, and from any pair of distinct cities, either exactly 0 or 2 other cities can be reached from both cities by a single road. Compute the number of ways HMMTLand could have constructed the roads. | 875 | Let the cities be numbered $1,2,3,4,5,6,7,8$. WLOG, 1 is connected to 2,3 , and 4 . First suppose 2 and 3 are connected; then 3 and 1 share a second common neighbor, which must be 4 (as 1 is not connected to anything else). Likewise 2 and 4 are connected, and so 5, 6, 7, 8 are pairwise connected as well, so the graph consists of two disjoint copies of $K_{4}$ : There are $\frac{1}{2}\binom{8}{4}=35$ ways to partition the 8 vertices into two groups of 4 , so there are 35 such graphs. Otherwise, none of 2,3,4 are connected to each other. Then 2 and 3 must share a common neighbor, as must 3 and 4 , and 2 and 4 . If these are the same neighbor, this vertex would share all three neighbors with 1, so they must be pairwise distinct. The last vertex must then be connected to these three, creating a cube graph. A cube has 48 symmetries, so the number of such graphs is $\frac{8!}{48}=840$. The total is $35+840=875$. |
3788 | 2 | 6 | 8 | We have a polyhedron such that an ant can walk from one vertex to another, traveling only along edges, and traversing every edge exactly once. What is the smallest possible total number of vertices, edges, and faces of this polyhedron? | 20 | This is obtainable by construction. Consider two tetrahedrons glued along a face; this gives us 5 vertices, 9 edges, and 6 faces, for a total of 20 , and one readily checks that the required Eulerian path exists. Now, to see that we cannot do better, first notice that the number $v$ of vertices is at least 5 , since otherwise we must have a tetrahedron, which does not have an Eulerian path. Each vertex is incident to at least 3 edges, and in fact, since there is an Eulerian path, all except possibly two vertices are incident to an even number of edges. So the number of edges is at least $(3+3+4+4+4) / 2$ (since each edge meets two vertices) $=9$. Finally, if $f=4$ then each face must be a triangle, because there are only 3 other faces for it to share edges with, and we are again in the case of a tetrahedron, which is impossible; therefore $f \geq 5$. So $f+v+e \geq 5+5+9=19$. But since $f+v-e=2-2 g$ (where $g$ is the number of holes in the polyhedron), $f+v+e$ must be even. This strengthens our bound to 20 as needed. |
3794 | 1 | 7 | 8 | Anne-Marie has a deck of 16 cards, each with a distinct positive factor of 2002 written on it. She shuffles the deck and begins to draw cards from the deck without replacement. She stops when there exists a nonempty subset of the cards in her hand whose numbers multiply to a perfect square. What is the expected number of cards in her hand when she stops? | \frac{837}{208} | Note that $2002=2 \cdot 7 \cdot 11 \cdot 13$, so that each positive factor of 2002 is included on exactly one card. Each card can identified simply by whether or not it is divisible by each of the 4 primes, and we can uniquely achieve all of the $2^{4}$ possibilities. Also, when considering the product of the values on many cards, we only care about the values of the exponents in the prime factorization modulo 2, as we have a perfect square exactly when each exponent is even. Now suppose Anne-Marie has already drawn $k$ cards. Then there are $2^{k}$ possible subsets of cards from those she has already drawn. Note that if any two of these subsets have products with the same four exponents modulo 2, then taking the symmetric difference yields a subset of cards in her hand where all four exponents are $0(\bmod 2)$, which would cause her to stop. Now when she draws the $(k+1)$th card, she achieves a perfect square subset exactly when the exponents modulo 2 match those from a subset of the cards she already has. Thus if she has already drawn $k$ cards, she will not stop if she draws one of $16-2^{k}$ cards that don't match a subset she already has. Let $p_{k}$ be the probability that Anne-Marie draws at least $k$ cards. We have the recurrence $$p_{k+2}=\frac{16-2^{k}}{16-k} p_{k+1}$$ because in order to draw $k+2$ cards, the $(k+1)$th card, which is drawn from the remaining $16-k$ cards, must not be one of the $16-2^{k}$ cards that match a subset of Anne-Marie's first $k$ cards. We now compute $$\begin{aligned} & p_{1}=1 \\ & p_{2}=\frac{15}{16} \\ & p_{3}=\frac{14}{15} p_{2}=\frac{7}{8} \\ & p_{4}=\frac{12}{14} p_{3}=\frac{3}{4} \\ & p_{5}=\frac{8}{13} p_{4}=\frac{6}{13} \\ & p_{6}=0 \end{aligned}$$ The expected number of cards that Anne-Marie draws is $$p_{1}+p_{2}+p_{3}+p_{4}+p_{5}=1+\frac{15}{16}+\frac{7}{8}+\frac{3}{4}+\frac{6}{13}=\frac{837}{208}$$ |
3804 | 1 | 7 | 8 | Define the sequence $a_{1}, a_{2} \ldots$ as follows: $a_{1}=1$ and for every $n \geq 2$, $a_{n}= \begin{cases}n-2 & \text { if } a_{n-1}=0 \\ a_{n-1}-1 & \text { if } a_{n-1} \neq 0\end{cases}$. A non-negative integer $d$ is said to be jet-lagged if there are non-negative integers $r, s$ and a positive integer $n$ such that $d=r+s$ and that $a_{n+r}=a_{n}+s$. How many integers in $\{1,2, \ldots, 2016\}$ are jet-lagged? | 51 | Let $N=n+r$, and $M=n$. Then $r=N-M$, and $s=a_{N}-a_{M}$, and $d=r+s=\left(a_{N}+N\right)-\left(a_{M}+M\right)$. So we are trying to find the number of possible values of $\left(a_{N}+N\right)-\left(a_{M}+M\right)$, subject to $N \geq M$ and $a_{N} \geq a_{M}$. Divide the $a_{i}$ into the following "blocks": - $a_{1}=1, a_{2}=0$ - $a_{3}=1, a_{4}=0$ - $a_{5}=3, a_{6}=2, a_{7}=1, a_{8}=0$ - $a_{9}=7, a_{10}=6, \ldots, a_{16}=0$ and so on. The $k^{t h}$ block contains $a_{i}$ for $2^{k-1}<i \leq 2^{k}$. It's easy to see by induction that $a_{2^{k}}=0$ and thus $a_{2^{k}+1}=2^{k}-1$ for all $k \geq 1$. Within each block, the value $a_{n}+n$ is constant, and for the $k$ th block $(k \geq 1)$ it equals $2^{k}$. Therefore, $d=\left(a_{N}+N\right)-\left(a_{M}+M\right)$ is the difference of two powers of 2 , say $2^{n}-2^{m}$. For any $n \geq 1$, it is clear there exists an $N$ such that $a_{N}+N=2^{n}$ (consider the $n^{\text {th }}$ block). We can guarantee $a_{N} \geq a_{M}$ by setting $M=2^{m}$. Therefore, we are searching for the number of integers between 1 and 2016 that can be written as $2^{n}-2^{m}$ with $n \geq m \geq 1$. The pairs $(n, m)$ with $n>m \geq 1$ and $n \leq 10$ all satisfy $1 \leq 2^{n}-2^{m} \leq 2016$ (45 possibilities). In the case that $n=11$, we have that $2^{n}-2^{m} \leq 2016$ so $2^{m} \geq 32$, so $m \geq 5$ (6 possibilities). There are therefore $45+6=51$ jetlagged numbers between 1 and 2016. |
3809 | 3 | 5 | 8 | Max repeatedly throws a fair coin in a hurricane. For each throw, there is a $4 \%$ chance that the coin gets blown away. He records the number of heads $H$ and the number of tails $T$ before the coin is lost. (If the coin is blown away on a toss, no result is recorded for that toss.) What is the expected value of $|H-T|$? | \frac{24}{7} | In all solutions, $p=\frac{1}{25}$ will denote the probability that the coin is blown away. Let $D=|H-T|$. Note that if $D \neq 0$, the expected value of $D$ is not changed by a coin flip, whereas if $D=0$, the expected value of $D$ increases by 1. Therefore $\mathbf{E}(D)$ can be computed as the sum over all $n$ of the probability that the $n$th coin flip occurs when $D=0$. This only occurs when $n=2 k+1$ is odd, where the probability that the first $n$ coin flips occur is $(1-p)^{2 k+1}$ and the probability that $D=0$ after the first $n-1$ flips is $\frac{\binom{2 k}{k}}{4^{k}}$. Therefore $$\begin{aligned} \mathbf{E}(D) & =(1-p) \sum_{k=0}^{\infty}\left(\frac{1-p}{2}\right)^{2 k}\binom{2 k}{k} \\ & =\frac{1-p}{\sqrt{1-(1-p)^{2}}} \end{aligned}$$ using the generating function $$\sum_{k=0}^{\infty}\binom{2 k}{k} x^{k}=\frac{1}{\sqrt{1-4 x}}$$ Plugging in $p=\frac{1}{25}$ yields $\mathbf{E}(D)=\frac{24}{7}$. |
3818 | 2 | 6 | 8 | Somewhere in the universe, $n$ students are taking a 10-question math competition. Their collective performance is called laughable if, for some pair of questions, there exist 57 students such that either all of them answered both questions correctly or none of them answered both questions correctly. Compute the smallest $n$ such that the performance is necessarily laughable. | 253 | Let $c_{i, j}$ denote the number of students correctly answering questions $i$ and $j(1 \leq i<j \leq 10)$, and let $w_{i, j}$ denote the number of students getting both questions wrong. An individual student answers $k$ questions correctly and $10-k$ questions incorrectly. This student answers $\binom{k}{2}$ pairs of questions correctly and $\binom{10-k}{2}$ pairs of questions incorrectly. Now observe that $$\binom{k}{2}+\binom{10-k}{2}=k^{2}-10 k+45=(k-5)^{2}+20 \geq 20$$ Therefore, $$\sum_{1 \leq i<j \leq 10} c_{i, j}+w_{i, j} \geq 20 n$$ Now if the performance is not laughable, then $c_{i, j} \leq 56$ and $w_{i, j} \leq 56$ for all $1 \leq i<j \leq 10$. Observe that there are $2\binom{10}{2}=90$ of these variables. Hence, in a boring performance, $$20 n \leq \sum_{1 \leq i<j \leq 10} c_{i, j}+w_{i, j} \leq 90 \cdot 56=5040$$ or $n \leq 252$. In particular this implies that if $n \geq 253$, the performance is laughable. This is the best bound because $\binom{10}{5}=252$, and if each of 252 students correctly answers a different 5 element subset of the 10 questions, then $c_{i, j}=w_{i, j}=56$ for all $1 \leq i<j \leq 10$. |
3819 | 2 | 6 | 8 | Find all real numbers $k$ such that $r^{4}+k r^{3}+r^{2}+4 k r+16=0$ is true for exactly one real number $r$. | \pm \frac{9}{4} | Any real quartic has an even number of real roots with multiplicity, so there exists real $r$ such that $x^{4}+k x^{3}+x^{2}+4 k x+16$ either takes the form $(x+r)^{4}$ (clearly impossible) or $(x+r)^{2}\left(x^{2}+a x+b\right)$ for some real $a, b$ with $a^{2}<4 b$. Clearly $r \neq 0$, so $b=\frac{16}{r^{2}}$ and $4 k=4(k)$ yields $\frac{32}{r}+a r^{2}=4(2 r+a) \Longrightarrow a\left(r^{2}-4\right)=8 \frac{r^{2}-4}{r}$. Yet $a \neq \frac{8}{r}$ (or else $a^{2}=4 b$ ), so $r^{2}=4$, and $1=r^{2}+2 r a+\frac{16}{r^{2}} \Longrightarrow a=\frac{-7}{2 r}$. Thus $k=2 r-\frac{7}{2 r}= \pm \frac{9}{4}$ (since $r= \pm 2$ ). |
3821 | 2 | 6 | 8 | Reimu and Sanae play a game using 4 fair coins. Initially both sides of each coin are white. Starting with Reimu, they take turns to color one of the white sides either red or green. After all sides are colored, the 4 coins are tossed. If there are more red sides showing up, then Reimu wins, and if there are more green sides showing up, then Sanae wins. However, if there is an equal number of red sides and green sides, then neither of them wins. Given that both of them play optimally to maximize the probability of winning, what is the probability that Reimu wins? | \frac{5}{16} | Clearly Reimu will always color a side red and Sanae will always color a side green, because their situation is never worse off when a side of a coin changes to their own color. Since the number of red-only coins is always equal to the number of green-only coins, no matter how Reimu and Sanae color the coins, they will have an equal probability of winning by symmetry, so instead they will cooperate to make sure that the probability of a tie is minimized, which is when all 4 coins have different colors on both sides (which can easily be achieved by Reimu coloring one side of a new coin red and Sanae immediately coloring the opposite side green). Therefore, the probability of Reimu winning is $\frac{\binom{4}{3}+\binom{4}{4}}{2^{4}}=\frac{5}{16}$. |
3832 | 4 | 4 | 8 | Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 black balls. An urn is randomly selected, and then a ball inside of that urn is removed. We then repeat the process of selecting an urn and drawing out a ball, without returning the first ball. What is the probability that the first ball drawn was red, given that the second ball drawn was black? | 7/15 | This is a case of conditional probability; the answer is the probability that the first ball is red and the second ball is black, divided by the probability that the second ball is black. First, we compute the numerator. If the first ball is drawn from Urn A, we have a probability of $2 / 6$ of getting a red ball, then a probability of $1 / 2$ of drawing the second ball from Urn B, and a further probability of $3 / 6$ of drawing a black ball. If the first ball is drawn from Urn B, we have probability $3 / 6$ of getting a red ball, then $1 / 2$ of drawing the second ball from Urn B, and $3 / 5$ of getting a black ball. So our numerator is $$ \frac{1}{2}\left(\frac{2}{6} \cdot \frac{1}{2} \cdot \frac{3}{6}+\frac{3}{6} \cdot \frac{1}{2} \cdot \frac{3}{5}\right)=\frac{7}{60} $$ We similarly compute the denominator: if the first ball is drawn from Urn A, we have a probability of $1 / 2$ of drawing the second ball from Urn B, and $3 / 6$ of drawing a black ball. If the first ball is drawn from Urn B, then we have probability $3 / 6$ that it is red, in which case the second ball will be black with probability $(1 / 2) \cdot(3 / 5)$, and probability $3 / 6$ that the first ball is black, in which case the second is black with probability $(1 / 2) \cdot(2 / 5)$. So overall, our denominator is $$ \frac{1}{2}\left(\frac{1}{2} \cdot \frac{3}{6}+\frac{3}{6}\left[\frac{1}{2} \cdot \frac{3}{5}+\frac{1}{2} \cdot \frac{2}{5}\right]\right)=\frac{1}{4} $$ Thus, the desired conditional probability is $(7 / 60) /(1 / 4)=7 / 15$. |
3833 | 1 | 7 | 8 | An ant starts out at $(0,0)$. Each second, if it is currently at the square $(x, y)$, it can move to $(x-1, y-1),(x-1, y+1),(x+1, y-1)$, or $(x+1, y+1)$. In how many ways can it end up at $(2010,2010)$ after 4020 seconds? | $\binom{4020}{1005}^{2}$ | Note that each of the coordinates either increases or decreases the x and y coordinates by 1. In order to reach 2010 after 4020 steps, each of the coordinates must be increased 3015 times and decreased 1005 times. A permutation of 3015 plusses and 1005 minuses for each of $x$ and $y$ uniquely corresponds to a path the ant could take to $(2010,2010)$, because we can take ordered pairs from the two lists and match them up to a valid step the ant can take. So the number of ways the ant can end up at $(2010,2010)$ after 4020 seconds is equal to the number of ways to arrange plusses and minuses for both $x$ and $y$, or $\left(\binom{4020}{1005}\right)^{2}$. |
3858 | 3 | 5 | 8 | Suppose $E, I, L, V$ are (not necessarily distinct) nonzero digits in base ten for which the four-digit number $\underline{E} \underline{V} \underline{I} \underline{L}$ is divisible by 73 , and the four-digit number $\underline{V} \underline{I} \underline{L} \underline{E}$ is divisible by 74 . Compute the four-digit number $\underline{L} \underline{I} \underline{V} \underline{E}$. | 9954 | Let $\underline{E}=2 k$ and $\underline{V} \underline{I} \underline{L}=n$. Then $n \equiv-2000 k(\bmod 73)$ and $n \equiv-k / 5(\bmod 37)$, so $n \equiv 1650 k(\bmod 2701)$. We can now exhaustively list the possible cases for $k$ : - if $k=1$, then $n \equiv 1650$ which is not possible; - if $k=2$, then $n \equiv 2 \cdot 1650 \equiv 599$, which gives $E=4$ and $n=599$; - if $k=3$, then $n \equiv 599+1650 \equiv 2249$ which is not possible; - if $k=4$, then $n \equiv 2249+1650 \equiv 1198$ which is not possible. Hence, we must have $(E, V, I, L)=(4,5,9,9)$, so $\underline{L} \underline{I} \underline{V} \underline{E}=9954$. |
3863 | 2 | 6 | 8 | A point $P$ lies at the center of square $A B C D$. A sequence of points $\left\{P_{n}\right\}$ is determined by $P_{0}=P$, and given point $P_{i}$, point $P_{i+1}$ is obtained by reflecting $P_{i}$ over one of the four lines $A B, B C, C D, D A$, chosen uniformly at random and independently for each $i$. What is the probability that $P_{8}=P$ ? | \frac{1225}{16384} | Solution 1. WLOG, $A B$ and $C D$ are horizontal line segments and $B C$ and $D A$ are vertical. Then observe that we can consider the reflections over vertical lines separately from those over horizontal lines, as each reflection over a vertical line moves $P_{i}$ horizontally to point $P_{i+1}$, and vice versa. Now consider only the reflections over horizontal segments $A B$ and $C D$. Note that it is impossible for $P_{8}$ to be in the same location vertical location as $P$ if there are an odd number of these reflections. Then we consider the reflections in pairs: let $w$ denote reflecting twice over $A B$, let $x$ denote reflecting over $A B$ and then $C D$, let $y$ denote reflecting over $C D$ and then $A B$, and let $z$ denote reflecting twice over $C D$. Note that both $w$ and $z$ preserve the position of our point. Also note that in order to end at the same vertical location as $P$, we must have an equal number of $x$ 's and $y$ 's. Now we count the number of sequences of length at most 4 with this property: - Case 1: Length 0 There is just the empty sequence here, so 1 . - Case 2: Length 1 There are just the sequences $w$ and $z$, so 2 . - Case 3: Length 2 We may either have an $x$ and a $y$ or two characters that are either $w$ or $z$. There are 2 sequences of the former type and 4 of the latter, for 6 total. - Case 4: Length 3 There are 12 sequences with an $x$, a $y$, and either a $w$ or a $z$, and 8 sequences of only $w$ 's and $z$ 's, for 12 total. - Case 5: Length 4 There are 6 sequences of $2 x$ 's and $2 y$ 's, 48 with one of each and two terms that are either $w$ or $z$, and 16 of just $w$ 's and $z$ 's, for a total of 70 . Now let the number of such sequences of length $k$ be $a_{k}$ (so $a_{3}=20$ ). Note that these counts work also if we consider only reflections over vertical line segments $B C$ and $A D$. Now to finish, we only need to count the number of ways to combine 2 valid sequences of total length 4 . This is $$\sum_{i=0}^{4} a_{i} a_{4-i}\binom{8}{2 i}$$ as there are $a_{i}$ sequences of reflections over $A B$ and $C D, a_{4-i}$ sequences of reflections over $B C$ and $A D$ such that there are 8 total reflections, and $\binom{8}{2 i}$ ways to choose which of the 8 reflections will be over $A B$ or $C D$. We compute that this sum is $1 \cdot 70 \cdot 1+2 \cdot 20 \cdot 28+6 \cdot 6 \cdot 70+20 \cdot 2 \cdot 28+70 \cdot 1 \cdot 1=4900$ total sequences of reflections that place $P_{8}$ at $P$. There are of course $4^{8}=65536$ total sequences of 8 reflections, each chosen uniformly at random, so our answer is $\frac{4900}{65536}=\frac{1225}{16384}$. |
3864 | 3 | 5 | 8 | Bobbo starts swimming at 2 feet/s across a 100 foot wide river with a current of 5 feet/s. Bobbo doesn't know that there is a waterfall 175 feet from where he entered the river. He realizes his predicament midway across the river. What is the minimum speed that Bobbo must increase to make it to the other side of the river safely? | 3 \text{ feet/s} | When Bobbo is midway across the river, he has travelled 50 feet. Going at a speed of 2 feet/s, this means that Bobbo has already been in the river for $\frac{50 \text{ feet}}{2 \text{ feet/s}} = 25 \text{ s}$. Then he has traveled 5 feet/s $\cdot$ 25 s = 125 feet down the river. Then he has 175 feet - 125 feet = 50 feet left to travel downstream before he hits the waterfall. Bobbo travels at a rate of 5 feet/s downstream. Thus there are $\frac{50 \text{ feet}}{5 \text{ feet/s}} = 10 \text{ s}$ before he hits the waterfall. He still has to travel 50 feet horizontally across the river. Thus he must travel at a speed of $\frac{50 \text{ feet}}{10 \text{ s}} = 5$ feet/s. This is a 3 feet/s difference from Bobbo's original speed of 2 feet/s. |
3891 | 1 | 7 | 8 | Compute the number of sequences of integers $(a_{1}, \ldots, a_{200})$ such that the following conditions hold. - $0 \leq a_{1}<a_{2}<\cdots<a_{200} \leq 202$. - There exists a positive integer $N$ with the following property: for every index $i \in\{1, \ldots, 200\}$ there exists an index $j \in\{1, \ldots, 200\}$ such that $a_{i}+a_{j}-N$ is divisible by 203. | 20503 | Let $m:=203$ be an integer not divisible by 3. We'll show the answer for general such $m$ is $m\left\lceil\frac{m-1}{2}\right\rceil$. Let $x, y, z$ be the three excluded residues. Then $N$ works if and only if $\{x, y, z\} \equiv\{N-x, N-y, N-z\} (\bmod m)$. Since $x, y, z(\bmod m)$ has opposite orientation as $N-x, N-y, N-z(\bmod m)$, this is equivalent to $x, y, z$ forming an arithmetic progression (in some order) modulo $m$ centered at one of $x, y, z$ (or algebraically, one of $N \equiv 2 x \equiv y+z, N \equiv 2 y \equiv z+x, N \equiv 2 z \equiv x+y$ holds, respectively). Since $3 \nmid m$, it's impossible for more than one of these congruences to hold. So the number of distinct 3-sets corresponding to arithmetic progressions is $m\left\lceil\frac{m-1}{2}\right\rceil$. Since our specific $m=203$ is odd this gives $m \frac{m-1}{2}=203 \cdot 101=20503$. |
3897 | 4 | 4 | 8 | If $a, b, c>0$, what is the smallest possible value of $\left\lfloor\frac{a+b}{c}\right\rfloor+\left\lfloor\frac{b+c}{a}\right\rfloor+\left\lfloor\frac{c+a}{b}\right\rfloor$? (Note that $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.) | 4 | Since $\lfloor x\rfloor>x-1$ for all $x$, we have that $$\begin{aligned} \left\lfloor\frac{a+b}{c}\right\rfloor+\left\lfloor\frac{b+c}{a}\right\rfloor+\left\lfloor\frac{c+a}{b}\right\rfloor & >\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}-3 \\ & =\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)-3 \end{aligned}$$ But by the AM-GM inequality, each of the first three terms in the last line is at least 2. Therefore, the lefthand side is greater than $2+2+2-3=3$. Since it is an integer, the smallest value it can be is 4. This is in fact attainable by letting $(a, b, c)=(6,8,9)$. |
3915 | 3 | 5 | 8 | A sequence of positive integers is defined by $a_{0}=1$ and $a_{n+1}=a_{n}^{2}+1$ for each $n \geq 0$. Find $\operatorname{gcd}(a_{999}, a_{2004})$. | 677 | If $d$ is the relevant greatest common divisor, then $a_{1000}=a_{999}^{2}+1 \equiv 1=a_{0}(\bmod d)$, which implies (by induction) that the sequence is periodic modulo $d$, with period 1000 . In particular, $a_{4} \equiv a_{2004} \equiv 0$. So $d$ must divide $a_{4}$. Conversely, we can see that $a_{5}=a_{4}^{2}+1 \equiv 1=a_{0}$ modulo $a_{4}$, so (again by induction) the sequence is periodic modulo $a_{4}$ with period 5 , and hence $a_{999}, a_{2004}$ are indeed both divisible by $a_{4}$. So the answer is $a_{4}$, which we can compute directly; it is 677. |
3921 | 2 | 6 | 8 | A classroom consists of a $5 \times 5$ array of desks, to be filled by anywhere from 0 to 25 students, inclusive. No student will sit at a desk unless either all other desks in its row or all others in its column are filled (or both). Considering only the set of desks that are occupied (and not which student sits at each desk), how many possible arrangements are there? | 962 | The set of empty desks must be of the form (non-full rows) $\times$ (non-full columns): each empty desk is in a non-full column and a non-full row, and the given condition implies that each desk in such a position is empty. So if there are fewer than 25 students, then both of these sets are nonempty; we have $2^{5}-1=31$ possible sets of non-full rows, and 31 sets of non-full columns, for 961 possible arrangements. Alternatively, there may be 25 students, and then only 1 arrangement is possible. Thus there are 962 possibilities altogether. |
3927 | 1 | 7 | 8 | A frog is at the point $(0,0)$. Every second, he can jump one unit either up or right. He can only move to points $(x, y)$ where $x$ and $y$ are not both odd. How many ways can he get to the point $(8,14)$? | 330 | When the frog is at a point $(x, y)$ where $x$ and $y$ are both even, then if that frog chooses to move right, his next move will also have to be a step right; similarly, if he moves up, his next move will have to be up. If we 'collapse' each double step into one step, the problem simply becomes how many ways are there to move to the point $(4,7)$ using only right and up steps, with no other restrictions. That is 11 steps total, so the answer is $\binom{11}{4}=330$. |
3930 | 1 | 7 | 8 | A unit square $A B C D$ and a circle $\Gamma$ have the following property: if $P$ is a point in the plane not contained in the interior of $\Gamma$, then $\min (\angle A P B, \angle B P C, \angle C P D, \angle D P A) \leq 60^{\circ}$. The minimum possible area of $\Gamma$ can be expressed as $\frac{a \pi}{b}$ for relatively prime positive integers $a$ and $b$. Compute $100 a+b$. | 106 | Note that the condition for $\Gamma$ in the problem is equivalent to the following condition: if $\min (\angle A P B, \angle B P C, \angle C P D, \angle D P A)>60^{\circ}$, then $P$ is contained in the interior of $\Gamma$. Let $X_{1}, X_{2}, X_{3}$, and $X_{4}$ be the four points in $A B C D$ such that $A B X_{1}, B C X_{2}, C D X_{3}$, and $D A X_{4}$ are all equilateral triangles. Now, let $\Omega_{1}, \Omega_{2}, \Omega_{3}$, and $\Omega_{4}$ be the respective circumcircles of these triangles, and let the centers of these circles be $O_{1}, O_{2}, O_{3}$, and $O_{4}$. Note that the set of points $P$ such that $\angle A P B, \angle B P C, \angle C P D, \angle D P A>60^{\circ}$ is the intersection of $\Omega_{1}, \Omega_{2}, \Omega_{3}$, and $\Omega_{4}$. We want to find the area of the minimum circle containing this intersection. Let $\Gamma_{1}$ and $\Gamma_{2}$ intersect at $B$ and $B^{\prime}$. Define $C^{\prime}, D^{\prime}$ and $A^{\prime}$ similarly. It is not hard to see that the circumcircle of square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is the desired circle. Now observe that $\angle A B^{\prime} D^{\prime}=\angle A B^{\prime} D=60^{\circ}$. Similarly, $\angle A D^{\prime} B^{\prime}=60^{\circ}$, so $A B^{\prime} D^{\prime}$ is equilateral. Its height is the distance from $A$ to $B^{\prime} D^{\prime}$, which is $\frac{1}{\sqrt{2}}$, so its side length is $\frac{\sqrt{6}}{3}$. This is also the diameter of the desired circle, so its area is $\frac{\pi}{4} \cdot \frac{6}{9}=\frac{\pi}{6}$. |
3934 | 1 | 7 | 8 | Three distinct vertices are randomly selected among the five vertices of a regular pentagon. Let $p$ be the probability that the triangle formed by the chosen vertices is acute. Compute $10 p$. | 5 | The only way for the three vertices to form an acute triangle is if they consist of two adjacent vertices and the vertex opposite their side. Since there are 5 ways to choose this and $\binom{5}{3}=10$ ways to choose the three vertices, we have $p=\frac{5}{10}=\frac{1}{2}$. |
3936 | 1 | 7 | 8 | Compute the remainder when 10002000400080016003200640128025605121024204840968192 is divided by 100020004000800160032. | 40968192 | Let $X_{k}$ denote $2^{k}$ except with leading zeroes added to make it four digits long. Let $\overline{a b c \cdots}$ denote the number obtained upon concatenating $a, b, c, \ldots$ We have $$2^{6} \cdot \overline{X_{0} X_{1} \ldots X_{5}}=\overline{X_{6} X_{7} \ldots X_{11}}$$ Therefore, $\overline{X_{0} X_{1} \ldots X_{5}}$ divides $\overline{X_{0} X_{1} \ldots X_{11}}$, meaning the remainder when $\overline{X_{0} X_{1} \ldots X_{13}}$ is divided by $\overline{X_{0} X_{1} \ldots X_{5}}$ is $$\overline{X_{12} X_{13}}=40968192$$ |
3938 | 1 | 7 | 8 | Compute the number of permutations $\pi$ of the set $\{1,2, \ldots, 10\}$ so that for all (not necessarily distinct) $m, n \in\{1,2, \ldots, 10\}$ where $m+n$ is prime, $\pi(m)+\pi(n)$ is prime. | 4 | Since $\pi$ sends pairs $(m, n)$ with $m+n$ prime to pairs $\left(m^{\prime}, n^{\prime}\right)$ with $m^{\prime}+n^{\prime}$ prime, and there are only finitely many such pairs, we conclude that if $m+n$ is composite, then so is $\pi(m)+\pi(n)$. Also note that $2 \pi(1)=\pi(1)+\pi(1)$ is prime because $2=1+1$ is prime. Thus, $\pi(1)=1$. Now, since $1+2,1+4,1+6$, and $1+10$ are all prime, we know that $\pi(2), \pi(4), \pi(6)$, and $\pi(10)$ are all even. Additionally, since $8+2,8+6,8+6$, and $8+10$ are all composite, it is not hard to see that $\pi(8)$ must also be even. Therefore $\pi$ preserves parity. Now, draw a bipartite graph between the odd and even numbers where we have an edge between $a$ and $b$ if and only if $a+b$ composite. We now only need to compute automorphisms of this graph that fix 1. Note that the edges are precisely $1-8-7-2,3-6-9$, and $4-5-10$. Since 1 is a fixed point of $\pi$, we know that $\pi$ fixes $1,8,7$, and 2. Additionally, $\pi(6)=6$ and $\pi(5)=5$. We can swap 3 and 9, as well as 4 and 10. Thus, there are $2 \cdot 2=4$ possible permutations. |
3939 | 1 | 7 | 8 | For an integer $n$, let $f_{9}(n)$ denote the number of positive integers $d \leq 9$ dividing $n$. Suppose that $m$ is a positive integer and $b_{1}, b_{2}, \ldots, b_{m}$ are real numbers such that $f_{9}(n)=\sum_{j=1}^{m} b_{j} f_{9}(n-j)$ for all $n>m$. Find the smallest possible value of $m$. | 28 | Let $M=9$. Consider the generating function $$F(x)=\sum_{n \geq 1} f_{M}(n) x^{n}=\sum_{d=1}^{M} \sum_{k \geq 1} x^{d k}=\sum_{d=1}^{M} \frac{x^{d}}{1-x^{d}}$$ Observe that $f_{M}(n)=f_{M}\left(n+M\right.$ !) for all $n \geq 1$ (in fact, all $n \leq 0$ as well). Thus $f_{M}(n)$ satisfies a degree $m$ linear recurrence if and only if it eventually satisfies a degree $m$ linear recurrence. But the latter occurs if and only if $P(x) F(x)$ is a polynomial for some degree $m$ polynomial $P(x)$. Suppose $P(x) F(x)=Q(x)$ is a polynomial for some polynomial $P$ of degree $m$. We show that $x^{s}-1 \mid$ $P(x)$ for $s=1,2, \ldots, M$, or equivalently that $P(\omega)=0$ for all primitive $s$ th roots of unity $1 \leq s \leq M)$. Fix a primitive $s$ th root of unity $\omega$, and define a function $$F_{\omega}(z)=\left(1-\omega^{-1} z\right) \sum_{s \nmid d \leq M} \frac{z^{d}}{1-z^{d}}+\sum_{s \mid d \leq M} \frac{z^{d}}{1+\left(\omega^{-1} z\right)+\cdots+\left(\omega^{-1} z\right)^{d-1}}$$ for all $z$ where all denominators are nonzero (in particular, this includes $z=\omega$ ). Yet $F_{\omega}(z)-F(z)\left(1-\omega^{-1} z\right)=0$ for all complex $z$ such that $z^{1}, z^{2}, \ldots, z^{M} \neq 1$, so $P(z) F_{\omega}(z)-Q(z)(1-$ $\left.\omega^{-1} z\right)=0$ holds for all such $z$ as well. In particular, the rational function $P(x) F_{\omega}(x)-Q(x)\left(1-\omega^{-1} x\right)$ has infinitely many roots, so must be identically zero once we clear denominators. But no denominator vanishes at $x=\omega$, so we may plug in $x=\omega$ to the polynomial identity and then divide out by the original (nonzero) denominators to get $0=P(\omega) F_{\omega}(\omega)-Q(\omega)\left(1-\omega^{-1} \omega\right)=P(\omega) F_{\omega}(\omega)$. However, $$F_{\omega}(\omega)=\sum_{s \mid d \leq M} \frac{\omega^{d}}{1+\left(\omega^{-1} \omega\right)+\cdots+\left(\omega^{-1} \omega\right)^{d-1}}=\sum_{s \mid d \leq M} \frac{1}{d}$$ is a positive integer multiple of $1 / d$, and therefore nonzero. Thus $P(\omega)=0$, as desired. Conversely, if $x^{s}-1 \mid P(x)$ for $s=1,2, \ldots, M$, then $P(x)$ will clearly suffice. So we just want the degree of the least common multiple of the $x^{s}-1$ for $s=1,2, \ldots, M$, or just the number of roots of unity of order at most $M$, which is $\sum_{s=1}^{M} \phi(s)=1+1+2+2+4+2+6+4+6=28$. |
3949 | 4 | 4 | 8 | Determine the smallest positive integer $n \geq 3$ for which $$A \equiv 2^{10 n}\left(\bmod 2^{170}\right)$$ where $A$ denotes the result when the numbers $2^{10}, 2^{20}, \ldots, 2^{10 n}$ are written in decimal notation and concatenated (for example, if $n=2$ we have $A=10241048576$). | 14 | Note that $$2^{10 n}=1024^{n}=1.024^{n} \times 10^{3 n}$$ So $2^{10 n}$ has roughly $3 n+1$ digits for relatively small $n$'s. (Actually we have that for $0<x<1$, $$(1+x)^{2}=1+2 x+x^{2}<1+3 x$$ Therefore, $1.024^{2}<1.03^{2}<1.09,1.09^{2}<1.27,1.27^{2}<1.81<2$, and $2^{2}=4$, so $1.024^{16}<4$. Thus the conclusion holds for $n \leq 16$.) For any positive integer $n \leq 16$, $$A=\sum_{i=1}^{n} 2^{10 i} \times 10^{\sum_{j=i+1}^{n}(3 j+1)}$$ Let $$A_{i}=2^{10 i} \times 10^{\sum_{j=i+1}^{n}(3 j+1)}$$ for $1 \leq i \leq n$, then we know that $$A-2^{10 n}=\sum_{i=1}^{n-1} A_{i}$$ and $$A_{i}=2^{10 i+\sum_{j=i+1}^{n}(3 j+1)} \times 5^{\sum_{j=i+1}^{n}(3 j+1)}=2^{u_{i}} \times 5^{v_{i}}$$ where $u_{i}=10 i+\sum_{j=i+1}^{n}(3 j+1), v_{i}=\sum_{j=i+1}^{n}(3 j+1)$. We have that $$u_{i}-u_{i-1}=10-(3 i+1)=3(3-i)$$ Thus, for $1 \leq i \leq n-1, u_{i}$ is minimized when $i=1$ or $i=n-1$, with $u_{1}=\frac{3 n^{2}+5 n+12}{2}$ and $u_{n-1}=13 n-9$. When $n=5$, $$A-2^{10 n}=A_{1}+A_{2}+A_{3}+A_{4}=2^{10} \times 10^{46}+2^{20} \times 10^{39}+2^{30} \times 10^{29}+2^{40} \times 10^{16}$$ is at most divisible by $2^{57}$ instead of $2^{170}$. For all other $n$'s, we have that $u_{1} \neq u_{n-1}$, so we should have that both $170 \leq u_{1}$ and $170 \leq u_{n-1}$. Therefore, since $170 \leq u_{n-1}$, we have that $14 \leq n$. We can see that $u_{1}>170$ and $14<16$ in this case. Therefore, the minimum of $n$ is 14. |
3950 | 4 | 4 | 8 | Call a positive integer $N \geq 2$ "special" if for every $k$ such that $2 \leq k \leq N, N$ can be expressed as a sum of $k$ positive integers that are relatively prime to $N$ (although not necessarily relatively prime to each other). How many special integers are there less than $100$? | 50 | We claim that all odd numbers are special, and the only special even number is 2. For any even $N>2$, the numbers relatively prime to $N$ must be odd. When we consider $k=3$, we see that $N$ can't be expressed as a sum of 3 odd numbers. Now suppose that $N$ is odd, and we look at the binary decomposition of $N$, so write $N=2^{a_{1}}+2^{a_{2}}+\ldots+2^{a_{j}}$ as a sum of distinct powers of 2. Note that all these numbers only have factors of 2 and are therefore relatively prime to $N$. We see that $j<\log _{2} N+1$. We claim that for any $k \geq j$, we can write $N$ as a sum of $k$ powers of 2. Suppose that we have $N$ written as $N=2^{a_{1}}+2^{a_{2}}+\ldots+2^{a_{k}}$. Suppose we have at least one of these powers of 2 even, say $2^{a_{1}}$. We can then write $N=2^{a_{1}-1}+2^{a_{1}-1}+2^{a_{2}}+\ldots+2^{a_{k}}$, which is $k+1$ powers of 2. The only way this process cannot be carried out is if we write $N$ as a sum of ones, which corresponds to $k=N$. Therefore, this gives us all $k>\log _{2} N$. Now we consider the case $k=2$. Let $2^{a}$ be the largest power of 2 such that $2^{a}<N$. We can write $N=2^{a}+\left(N-2^{a}\right)$. Note that since $2^{a}$ and $N$ are relatively prime, so are $N-2^{a}$ and $N$. Note that $a<\log _{2} N$. Now similar to the previous argument, we can write $2^{a}$ as a sum of $k$ powers of 2 for $1<k<2^{a}$, and since $2^{a}>\frac{N}{2}$, we can achieve all $k$ such that $2 \leq k<\frac{N}{2}+1$. Putting these together, we see that since $\frac{N}{2}+1>\log _{2} N$ for $N \geq 3$, we can achieve all $k$ from 2 through $N$, where $N$ is odd. |
3953 | 1 | 7 | 8 | The area of the largest regular hexagon that can fit inside of a rectangle with side lengths 20 and 22 can be expressed as $a \sqrt{b}-c$, for positive integers $a, b$, and $c$, where $b$ is squarefree. Compute $100 a+10 b+c$. | 134610 | Let $s$ be the sidelength of the hexagon. We can view this problem as finding the maximal rectangle of with sides $s$ and $s \sqrt{3}$ that can fit inside this rectangle. Let $A B C D$ be a rectangle with $A B=20$ and $B C=22$ and let $X Y Z W$ be an inscribed rectangle with $X$ on $A B$ and $Y$ on $B C$ with $X Y=s$ and $Y Z=s \sqrt{3}$. Let $B X=a$ and $B Y=b$. Then, by similar triangles, we have $A X=b \sqrt{3}$ and $C Y=a \sqrt{3}$. Thus, we have $a+b \sqrt{3}=20$ and $a \sqrt{3}+b=22$. Solving gives us $a=11 \sqrt{3}-10$ and $b=10 \sqrt{3}-11$, so $s^{2}=a^{2}+b^{2}=884-440 \sqrt{3}$. Thus, the area of the hexagon is $\frac{s^{2} \cdot 3 \sqrt{3}}{2}=1326 \sqrt{3}-1980$. |
3961 | 1 | 7 | 8 | Suppose that $x$ and $y$ are complex numbers such that $x+y=1$ and that $x^{20}+y^{20}=20$. Find the sum of all possible values of $x^{2}+y^{2}$. | -90 | We have $x^{2}+y^{2}+2 x y=1$. Define $a=2 x y$ and $b=x^{2}+y^{2}$ for convenience. Then $a+b=1$ and $b-a=x^{2}+y^{2}-2 x y=(x-y)^{2}=2 b-1$ so that $x, y=\frac{\sqrt{2 b-1} \pm 1}{2}$. Then $x^{20}+y^{20}=\left(\frac{\sqrt{2 b-1}+1}{2}\right)^{20}+\left(\frac{\sqrt{2 b-1}-1}{2}\right)^{20}=\frac{1}{2^{20}}\left[(\sqrt{2 b-1}+1)^{20}+(\sqrt{2 b-1}-1)^{20}\right]=\frac{2}{2^{20}}\left[(\sqrt{2 b-1})^{20}+\binom{20}{2}(\sqrt{2 b-1})^{18}+\binom{20}{4}(\sqrt{2 b-1})^{16}+\ldots\right]=20$. We want to find the sum of distinct roots of the above polynomial in $b$; we first prove that the original polynomial is square-free. The conditions $x+y=1$ and $x^{20}+y^{20}=20$ imply that $x^{20}+(1-x)^{20}-20=0$; let $p(x)=x^{20}+(1-x)^{20}-20 . p$ is square-free if and only if $G C D\left(p, p^{\prime}\right)=c$ for some constant $c$: $G C D\left(p, p^{\prime}\right)=G C D\left(x^{20}+(1-x)^{20}-20,20\left(x^{19}-(1-x)^{19}\right)\right)=G C D\left(x^{20}-x(1-x)^{19}+(1-x)^{19}-20,20\left(x^{19}-(1-x)^{19}\right)\right)=G C D\left((1-x)^{19}-20, x^{19}-(1-x)^{19}\right)=G C D\left((1-x)^{19}-20, x^{19}-20\right)$. The roots of $x^{19}-20$ are $\sqrt[19]{20^{k}} \exp \left(\frac{2 \pi i k}{19}\right)$ for some $k=0,1, \ldots, 18$; the roots of $(1-x)^{19}-20$ are $1-\sqrt[19]{20^{k}} \exp \left(\frac{2 \pi i k}{19}\right)$ for some $k=0,1, \ldots, 18$. If $x^{19}-20$ and $(1-x)^{19}-20$ share a common root, then there exist integers $m, n$ such that $\sqrt[19]{20^{m}} \exp \left(\frac{2 \pi i m}{19}\right)=1-\sqrt[19]{20^{n}} \exp \left(\frac{2 \pi i n}{19}\right)$; since the imaginary parts of both sides must be the same, we have $m=n$ and $\sqrt[19]{20^{m}} \exp \left(\frac{2 \pi i m}{19}\right)=\frac{1}{2} \Longrightarrow 20^{m}=\frac{1}{2^{19}}$, a contradiction. Thus we have proved that the polynomial in $x$ has no double roots. Since for each $b$ there exists a unique pair $(x, y)$ (up to permutations) that satisfies $x^{2}+y^{2}=b$ and $(x+y)^{2}=1$, the polynomial in $b$ has no double roots. Let the coefficient of $b^{n}$ in the above equation be $\left[b^{n}\right]$. By Vieta's Formulas, the sum of all possible values of $b=x^{2}+y^{2}$ is equal to $-\frac{\left[b^{9}\right]}{\left[b^{10}\right]} . \quad\left[b^{10}\right]=\frac{2}{2^{20}}\left(2^{10}\right)$ and $\left[b^{9}\right]=\frac{2}{2^{20}}\left(-\binom{10}{1} 2^{9}+\binom{20}{2} 2^{9}\right)$; thus $-\frac{\left[b^{9}\right]}{\left[b^{10}\right]}=-\frac{\binom{10}{1} 2^{9}-\binom{20}{2} 2^{9}}{2^{10}}=-90$. |
3965 | 3 | 5 | 8 | Let $S_{0}=0$ and let $S_{k}$ equal $a_{1}+2 a_{2}+\ldots+k a_{k}$ for $k \geq 1$. Define $a_{i}$ to be 1 if $S_{i-1}<i$ and -1 if $S_{i-1} \geq i$. What is the largest $k \leq 2010$ such that $S_{k}=0$? | 1092 | Suppose that $S_{N}=0$ for some $N \geq 0$. Then $a_{N+1}=1$ because $N+1 \geq S_{N}$. The following table lists the values of $a_{k}$ and $S_{k}$ for a few $k \geq N$: $k$ & $a_{k}$ & $S_{k}$ \hline$N$ & & 0 \$N+1$ & 1 & $N+1$ \$N+2$ & 1 & $2 N+3$ \$N+3$ & -1 & $N$ \$N+4$ & 1 & $2 N+4$ \$N+5$ & -1 & $N-1$ \$N+6$ & 1 & $2 N+5$ \$N+7$ & -1 & $N-2$. We see inductively that, for every $i \geq 1$, $S_{N+2 i}=2 N+2+i$ and $S_{N+1+2 i}=N+1-i$ thus $S_{3 N+3}=0$ is the next $k$ for which $S_{k}=0$. The values of $k$ for which $S_{k}=0$ satisfy the recurrence relation $p_{n+1}=3 p_{n}+3$, and we compute that the first terms of the sequence are $0,3,12,39,120,363,1092$; hence 1092 is our answer. |
3966 | 1 | 7 | 8 | Let $A B C$ be an acute scalene triangle with circumcenter $O$ and centroid $G$. Given that $A G O$ is a right triangle, $A O=9$, and $B C=15$, let $S$ be the sum of all possible values for the area of triangle $A G O$. Compute $S^{2}$. | 288 | Note that we know that $O, H$, and $G$ are collinear and that $H G=2 O G$. Thus, let $O G=x$ and $H G=2 x$. We also have $\sin A=\frac{B C}{2 R}=\frac{5}{6}$, so $\cos A=\frac{\sqrt{11}}{6}$. Then, if $A G \perp O G$, then we have $x^{2}+A G^{2}=O G^{2}+A G^{2}=A O^{2}=81$ and $H G^{2}+A G^{2}=4 x^{2}+A G^{2}=A H^{2}=(2 R \cos A)^{2}=99$. Solving gives us $x=\sqrt{6}$ and $A G=5 \sqrt{3}$. Thus, the area of $A G O$ in this case is $\frac{1}{2} \cdot \sqrt{6} \cdot 5 \sqrt{3}=\frac{5 \sqrt{3}}{2}$. If we have $A O \perp O G$, then we have $99=A H^{2}=A O^{2}+O H^{2}=81+9 x^{2}$. This gives us $x=\sqrt{2}$. In this case, we have the area of $A G O$ is $\frac{1}{2} \cdot \sqrt{2} \cdot 9=\frac{9 \sqrt{2}}{2}$. Adding up the two areas gives us $S=12 \sqrt{2}$. Squaring gives $S^{2}=288$. |
3979 | 1 | 7 | 8 | Ten positive integers are arranged around a circle. Each number is one more than the greatest common divisor of its two neighbors. What is the sum of the ten numbers? | 28 | First note that all the integers must be at least 2, because the greatest common divisor of any two positive integers is at least 1. Let $n$ be the largest integer in the circle. The greatest common divisor of its two neighbors is $n-1$. Therefore, each of the two neighbors is at least $n-1$ but at most $n$, so since $n-1 \nmid n$ for $n-1 \geq 2$, they must both be equal to $n-1$. Let $m$ be one of the numbers on the other side of $n-1$ from $n$. Then $\operatorname{gcd}(n, m)=n-2$. Since $n-2 \geq 0, n-2 \mid n$ only for $n=3$ or 4. If $n=3$, each number must be 2 or 3, and it is easy to check that there is no solution. If $n=4$, then it is again not hard to find that there is a unique solution up to rotation, namely 4322343223. The only possible sum is therefore 28. |
3981 | 3 | 5 | 8 | Find the number of ordered pairs of integers $(a, b) \in\{1,2, \ldots, 35\}^{2}$ (not necessarily distinct) such that $a x+b$ is a "quadratic residue modulo $x^{2}+1$ and 35 ", i.e. there exists a polynomial $f(x)$ with integer coefficients such that either of the following equivalent conditions holds: - there exist polynomials $P, Q$ with integer coefficients such that $f(x)^{2}-(a x+b)=\left(x^{2}+1\right) P(x)+35 Q(x)$ - or more conceptually, the remainder when (the polynomial) $f(x)^{2}-(a x+b)$ is divided by (the polynomial) $x^{2}+1$ is a polynomial with (integer) coefficients all divisible by 35 . | 225 | By the Chinese remainder theorem, we want the product of the answers modulo 5 and modulo 7 (i.e. when 35 is replaced by 5 and 7, respectively). First we do the modulo 7 case. Since $x^{2}+1$ is irreducible modulo 7 (or more conceptually, in $\mathbb{F}_{7}[x]$ ), exactly half of the nonzero residues modulo $x^{2}+1$ and 7 (or just modulo $x^{2}+\overline{1}$ if we're working in $\mathbb{F}_{7}[x]$ ) are quadratic residues, i.e. our answer is $1+\frac{7^{2}-1}{2}=25$ (where we add back one for the zero polynomial). Now we do the modulo 5 case. Since $x^{2}+1$ factors as $(x+2)(x-2)$ modulo 5 (or more conceptually, in $\mathbb{F}_{5}[x]$ ), by the polynomial Chinese remainder theorem modulo $x^{2}+\overline{1}$ (working in $\mathbb{F}_{5}[x]$ ), we want the product of the number of polynomial quadratic residues modulo $x \pm \overline{2}$. By centering/evaluating polynomials at $\mp \overline{2}$ accordingly, the polynomial squares modulo these linear polynomials are just those reducing to integer squares modulo 5. So we have an answer of $\left(1+\frac{5-1}{2}\right)^{2}=9$ in this case. Our final answer is thus $25 \cdot 9=225$. |
3983 | 1 | 7 | 8 | A jar contains 8 red balls and 2 blue balls. Every minute, a ball is randomly removed. The probability that there exists a time during this process where there are more blue balls than red balls in the jar can be expressed as $\frac{a}{b}$ for relatively prime integers $a$ and $b$. Compute $100 a+b$. | 209 | One can show that the condition in the problem is satisfied if and only the last ball drawn is blue (which happens with probability $\frac{1}{5}$), or the blue balls are drawn second-to-last and third-to-last (which happens with probability $\frac{1}{\binom{10}{2}}=\frac{1}{45}$). Thus the total probability is $\frac{10}{45}=\frac{2}{9}$. |
3986 | 1 | 7 | 8 | Let $\pi$ be a uniformly random permutation of the set $\{1,2, \ldots, 100\}$. The probability that $\pi^{20}(20)=$ 20 and $\pi^{21}(21)=21$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100 a+b$. (Here, $\pi^{k}$ means $\pi$ iterated $k$ times.) | 1025 | We look at the cycles formed by $\pi$ Let $\operatorname{ord}_{\pi}(n)$ denote the smallest $m$ such that $\pi^{m}(n)=n$. In particular, the condition implies that $\operatorname{ord}_{\pi}(20) \mid 20$ and $\operatorname{ord}_{\pi}(21) \mid 21$. Claim 1. 20 and 21 cannot be in the same cycle. Proof. If 20 and 21 were in the same cycle, then $x=\operatorname{ord}_{\pi}(20)=\operatorname{ord}_{\pi}(21)$ for some $x$. Then $x>1$ since the cycle contains both 20 and 21, but $x|20, x| 21$ implies $x=1$, a contradiction. Claim 2. The probability that $a=\operatorname{ord}_{\pi}(20), b=\operatorname{ord}_{\pi}(21)$ for some fixed $a, b$ such that $a+b \leq 100$ is $\frac{1}{99 \cdot 100}$. Proof. We can just count these permutations. We first choose $a-1$ elements of $[100] \backslash\{20,21\}$ to be in the cycle of 20, then we similarly choose $b-1$ to be in the cycle of 21. We then have $(a-1)$! ways to reorder within the cycle of $20,(b-1)$! ways to reorder within the cycle of 21, and $(100-a-b)$! ways to permute the remaining elements. The total number of ways is just $$\frac{98!}{(a-1)!(b-1)!(100-a-b)!} \cdot(a-1)!(b-1)!(100-a-b)!=98!$$ so the probability this happens is just $\frac{98!}{100!}=\frac{1}{9900}$. Now, since $\operatorname{ord}_{\pi}(20) \mid 20$ and $\operatorname{ord}_{\pi}(21) \mid 21$, we have 6 possible values for $\operatorname{ord}_{\pi}(20)$ and 4 for $\operatorname{ord}_{\pi}(21)$, so in total we have a $\frac{6 \cdot 4}{9900}=\frac{2}{825}$ probability that the condition is satisfied. |
3987 | 3 | 5 | 8 | The Fibonacci sequence is defined as follows: $F_{0}=0, F_{1}=1$, and $F_{n}=F_{n-1}+F_{n-2}$ for all integers $n \geq 2$. Find the smallest positive integer $m$ such that $F_{m} \equiv 0(\bmod 127)$ and $F_{m+1} \equiv 1(\bmod 127)$. | 256 | First, note that 5 is not a quadratic residue modulo 127. We are looking for the period of the Fibonacci numbers $\bmod 127$. Let $p=127$. We work in $\mathbb{F}_{p^{2}}$ for the remainder of this proof. Let $\alpha$ and $\beta$ be the roots of $x^{2}-x-1$. Then we know that $F_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}$. Note that since $x \rightarrow x^{p}$ is an automorphism and since automorphisms cycle the roots of a polynomial we have that $\alpha^{p}=\beta$ and $\beta^{p}=\alpha$. Then $F_{p}=\frac{\alpha^{p}-\beta^{p}}{\alpha-\beta}=-1$ and $F_{p+1}=\frac{\alpha \beta-\beta \alpha}{\alpha-\beta}=0$ and similarly we obtain $F_{2 p+1}=1$ and $F_{2 p+2}=0$. Thus since $2 p+2$ is a power of 2 and since the period does not divide $p+1$, we must have the answer is $2 p+2=256$. |
3997 | 3 | 5 | 8 | Determine the largest real number $c$ such that for any 2017 real numbers $x_{1}, x_{2}, \ldots, x_{2017}$, the inequality $$\sum_{i=1}^{2016} x_{i}\left(x_{i}+x_{i+1}\right) \geq c \cdot x_{2017}^{2}$$ holds. | -\frac{1008}{2017} | Let $n=2016$. Define a sequence of real numbers \left\{p_{k}\right\} by $p_{1}=0$, and for all $k \geq 1$, $$p_{k+1}=\frac{1}{4\left(1-p_{k}\right)}$$ Note that, for every $i \geq 1$, $$\left(1-p_{i}\right) \cdot x_{i}^{2}+x_{i} x_{i+1}+p_{i+1} x_{i+1}^{2}=\left(\frac{x_{i}}{2 \sqrt{p_{i+1}}}+\sqrt{p_{i+1}} x_{i+1}\right)^{2} \geq 0$$ Summing from $i=1$ to $n$ gives $$\sum_{i=1}^{n} x_{i}\left(x_{i}+x_{i+1}\right) \geq-p_{n+1} x_{n+1}^{2}$$ One can show by induction that $p_{k}=\frac{k-1}{2 k}$. Therefore, our answer is $-p_{2017}=-\frac{1008}{2017}$. |
4000 | 3 | 5 | 8 | In how many ways can you rearrange the letters of "HMMTHMMT" such that the consecutive substring "HMMT" does not appear? | 361 | There are $8!/(4!2!2!)=420$ ways to order the letters. If the permuted letters contain "HMMT", there are $5 \cdot 4!/ 2!=60$ ways to order the other letters, so we subtract these. However, we have subtracted "HMMTHMMT" twice, so we add it back once to obtain 361 possibilities. |
4006 | 2 | 6 | 8 | The polynomial \( f(x)=x^{2007}+17 x^{2006}+1 \) has distinct zeroes \( r_{1}, \ldots, r_{2007} \). A polynomial \( P \) of degree 2007 has the property that \( P\left(r_{j}+\frac{1}{r_{j}}\right)=0 \) for \( j=1, \ldots, 2007 \). Determine the value of \( P(1) / P(-1) \). | 289/259 | For some constant \( k \), we have \( P(z)=k \prod_{j=1}^{2007}\left(z-\left(r_{j}+\frac{1}{r_{j}}\right)\right) \). Now writing \( \omega^{3}=1 \) with \( \omega \neq 1 \), we have \( \omega^{2}+\omega=-1 \). Then \( P(1) / P(-1)=\frac{k \prod_{j=1}^{2007}\left(1-\left(r_{j}+\frac{1}{r_{j}}\right)\right)}{k \prod_{j=1}^{2007}\left(-1-\left(r_{j}+\frac{1}{r_{j}}\right)\right)}=\prod_{j=1}^{2007} \frac{r_{j}^{2}-r_{j}+1}{r_{j}^{2}+r_{j}+1}=\prod_{j=1}^{2007} \frac{\left(-\omega-r_{j}\right)\left(-\omega^{2}-r_{j}\right)}{\left(\omega-r_{j}\right)\left(\omega^{2}-r_{j}\right)} =\frac{f(-\omega) f\left(-\omega^{2}\right)}{f(\omega) f\left(\omega^{2}\right)}=\frac{\left(-\omega^{2007}+17 \omega^{2006}+1\right)\left(-\left(\omega^{2}\right)^{2007}+17\left(\omega^{2}\right)^{2006}+1\right)}{\left(\omega^{2007}+17 \omega^{2006}+1\right)\left(\left(\omega^{2}\right)^{2007}+17\left(\omega^{2}\right)^{2006}+1\right)}=\frac{\left(17 \omega^{2}\right)(17 \omega)}{\left(2+17 \omega^{2}\right)(2+17 \omega)} =\frac{289}{4+34\left(\omega+\omega^{2}\right)+289}=\frac{289}{259} \). |
4018 | 2 | 6 | 8 | You are given a $10 \times 2$ grid of unit squares. Two different squares are adjacent if they share a side. How many ways can one mark exactly nine of the squares so that no two marked squares are adjacent? | 36 | Since each row has only two squares, it is impossible for two marked squares to be in the same row. Therefore, exactly nine of the ten rows contain marked squares. Consider two cases: Case 1: The first or last row is empty. These two cases are symmetrical, so assume without loss of generality that the first row is empty. There are two possibilities for the second row: either the first square is marked, or the second square is marked. Since the third row must contain a marked square, and it cannot be in the same column as the marked square in the second row, the third row is determined by the second. Similarly, all the remaining rows are determined. This leaves two possibilities if the first row is empty. Thus, there are four possibilities if the first or last row is empty. Case 2: The empty row is not the first or last. Then, there are two blocks of (one of more) consecutive rows of marked squares. As above, the configuration of the rows in each of the two blocks is determined by the position of the marked square in the first of its rows. That makes $2 \times 2=4$ possible configurations. There are eight possibilities for the empty row, making a total of 32 possibilities in this case. Together, there are 36 possible configurations of marked squares. |
4035 | 3 | 5 | 8 | Compute the number of positive real numbers $x$ that satisfy $\left(3 \cdot 2^{\left\lfloor\log _{2} x\right\rfloor}-x\right)^{16}=2022 x^{13}$. | 9 | Let $f(x)=3 \cdot 2^{\left\lfloor\log _{2} x\right\rfloor}-x$. Note that for each integer $i$, if $x \in\left[2^{i}, 2^{i+1}\right)$, then $f(x)=3 \cdot 2^{i}-x$. This is a line segment from $\left(2^{i}, 2^{i+1}\right)$ to $\left(2^{i+1}, 2^{i}\right)$, including the first endpoint but not the second. Now consider the function $f(x)^{16} / x^{13}$. This consists of segments of decreasing functions connecting $\left(2^{i}, 2^{3 i+16}\right)$ and $\left(2^{i+1}, 2^{3 i-13}\right)$. Note that for each $-1 \leq i \leq 7$, we have that $2^{3 i-13} \leq 2^{10}<2022<2^{11} \leq 2^{3 i+16}$. This gives us 9 solutions in total. |
4058 | 2 | 6 | 8 | Compute the sum of all 2-digit prime numbers $p$ such that there exists a prime number $q$ for which $100 q+p$ is a perfect square. | 179 | All squares must end with $0,1,4,5,6$, or 9, meaning that $p$ must end with 1 and 9. Moreover, since all odd squares are $1 \bmod 4$, we know that $p$ must be $1 \bmod 4$. This rules all primes except for $41,61,29,89$. Since $17^{2}=289,19^{2}=361,23^{2}=529,89,61$, and 29 all work. To finish, we claim that 41 does not work. If $100 q+41$ were a square, then since all odd squares are $1 \bmod 8$ we find that $4 q+1 \equiv 1(\bmod 8)$, implying that $q$ is even. But 241 is not a square, contradiction. The final answer is $29+61+89=179$. |
4060 | 2 | 6 | 8 | Define the sequence $\{x_{i}\}_{i \geq 0}$ by $x_{0}=2009$ and $x_{n}=-\frac{2009}{n} \sum_{k=0}^{n-1} x_{k}$ for all $n \geq 1$. Compute the value of $\sum_{n=0}^{2009} 2^{n} x_{n}$ | 2009 | We have $-\frac{n x_{n}}{2009}=x_{n-1}+x_{n-2}+\ldots+x_{0}=x_{n-1}+\frac{(n-1) x_{n-1}}{2009}$, which yields the recursion $x_{n}=\frac{n-2010}{n} x_{n-1}$. Unwinding this recursion, we find $x_{n}=(-1)^{n} \cdot 2009$. $\binom{2008}{n}$. Thus $\sum_{k=0}^{2009} 2^{n} x_{n} =\sum_{k=0}^{2009}(-2)^{n} \cdot 2009 \cdot\binom{2008}{n} =2009 \sum_{k=0}^{2008}(-2)^{n}\binom{2008}{n} =2009(-2+1)^{2008}$ as desired. |
4062 | 4 | 4 | 8 | You have infinitely many boxes, and you randomly put 3 balls into them. The boxes are labeled $1,2, \ldots$. Each ball has probability $1 / 2^{n}$ of being put into box $n$. The balls are placed independently of each other. What is the probability that some box will contain at least 2 balls? | 5 / 7 | Notice that the answer is the sum of the probabilities that boxes $1,2, \ldots$, respectively, contain at least 2 balls, since those events are mutually exclusive. For box $n$, the probability of having at least 2 balls is $$3\left[\left(1 / 2^{n}\right)^{2}\left(1-1 / 2^{n}\right)\right]+\left(1 / 2^{n}\right)^{3}=3 / 2^{2 n}-2 / 2^{3 n}=3 / 4^{n}-2 / 8^{n}$$ Summing to infinity using the geometric series formula, we get the answer (3/4)/(1$1 / 4)-(2 / 8) /(1-1 / 8)$, which is equal to $5 / 7$. |
4063 | 3 | 5 | 8 | The train schedule in Hummut is hopelessly unreliable. Train A will enter Intersection X from the west at a random time between 9:00 am and 2:30 pm; each moment in that interval is equally likely. Train B will enter the same intersection from the north at a random time between 9:30 am and 12:30 pm, independent of Train A; again, each moment in the interval is equally likely. If each train takes 45 minutes to clear the intersection, what is the probability of a collision today? | \frac{13}{48} | Suppose we fix the time at which Train B arrives at Intersection X; then call the interval during which Train A could arrive (given its schedule) and collide with Train B the 'disaster window.' We consider two cases: (i) Train B enters Intersection $X$ between 9:30 and 9:45. If Train B arrives at 9:30, the disaster window is from 9:00 to 10:15, an interval of $1 \frac{1}{4}$ hours. If Train B arrives at $9: 45$, the disaster window is $1 \frac{1}{2}$ hours long. Thus, the disaster window has an average length of $\left(1 \frac{1}{4}+1 \frac{1}{2}\right) \div 2=\frac{11}{8}$. From 9:00 to $2: 30$ is $5 \frac{1}{2}$ hours. The probability of a collision is thus $\frac{11}{8} \div 5 \frac{1}{2}=\frac{1}{4}$. (ii) Train B enters Intersection $X$ between 9:45 and 12:30. Here the disaster window is always $1 \frac{1}{2}$ hours long, so the probability of a collision is $1 \frac{1}{2} \div 5 \frac{1}{2}=\frac{3}{11}$. From 9:30 to $12: 30$ is 3 hours. Now case (i) occurs with probability $\frac{1}{4} \div 3=\frac{1}{12}$, and case (ii) occurs with probability $\frac{11}{12}$. The overall probability of a collision is therefore $\frac{1}{12} \cdot \frac{1}{4}+\frac{11}{12} \cdot \frac{3}{11}=\frac{1}{48}+\frac{1}{4}=\frac{13}{48}$. |
4085 | 1 | 7 | 8 | In triangle $A B C, \angle A B C$ is obtuse. Point $D$ lies on side $A C$ such that \angle A B D$ is right, and point $E$ lies on side $A C$ between $A$ and $D$ such that $B D$ bisects \angle E B C$. Find $C E$, given that $A C=35, B C=7$, and $B E=5$. | 10 | Reflect $A$ and $E$ over $B D$ to $A^{\prime}$ and $E^{\prime}$ respectively. Note that the angle conditions show that $A^{\prime}$ and $E^{\prime}$ lie on $A B$ and $B C$ respectively. $B$ is the midpoint of segment $A A^{\prime}$ and $C E^{\prime}=$ $B C-B E^{\prime}=2$. Menelaus' theorem now gives $$\frac{C D}{D A} \cdot \frac{A A^{\prime}}{A^{\prime} B} \cdot \frac{B E^{\prime}}{E^{\prime} C}=1$$ from which $D A=5 C D$ or $C D=A C / 6$. By the angle bisector theorem, $D E=5 C D / 7$, so that $C E=12 C D / 7=10$. |
4086 | 3 | 5 | 8 | Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved. | -\frac{1}{8} | The minimum value is $-\frac{1}{8}$. There are eight equality cases in total. The first one is $$\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right)$$ Cyclic shifting all the entries give three more quadruples. Moreover, flipping the sign $((a, b, c, d) \rightarrow (-a,-b,-c,-d))$ all four entries in each of the four quadruples give four more equality cases. Since the expression is cyclic, we could WLOG $a=\max \{a, b, c, d\}$. Let $$S(a, b, c, d)=(a-b)(b-c)(c-d)(d-a)$$ Note that we have given $(a, b, c, d)$ such that $S(a, b, c, d)=-\frac{1}{8}$. Therefore, to prove that $S(a, b, c, d) \geq -\frac{1}{8}$, we just need to consider the case where $S(a, b, c, d)<0$. - Exactly 1 of $a-b, b-c, c-d, d-a$ is negative. Since $a=\max \{a, b, c, d\}$, then we must have $d-a<0$. This forces $a>b>c>d$. Now, let us write $$S(a, b, c, d)=-(a-b)(b-c)(c-d)(a-d)$$ Write $a-b=y, b-c=x, c-d=w$ for some positive reals $w, x, y>0$. Plugging to the original condition, we have $$(d+w+x+y)^{2}+(d+w+x)^{2}+(d+w)^{2}+d^{2}-1=0(*)$$ and we want to prove that $w x y(w+x+y) \leq \frac{1}{8}$. Consider the expression $(*)$ as a quadratic in $d$ : $$4 d^{2}+d(6 w+4 x+2 y)+\left((w+x+y)^{2}+(w+x)^{2}+w^{2}-1\right)=0$$ Since $d$ is a real number, then the discriminant of the given equation has to be non-negative, i.e. we must have $$\begin{aligned} 4 & \geq 4\left((w+x+y)^{2}+(w+x)^{2}+w^{2}\right)-(3 w+2 x+y)^{2} \\ & =\left(3 w^{2}+2 w y+3 y^{2}\right)+4 x(w+x+y) \\ & \geq 8 w y+4 x(w+x+y) \\ & =4(x(w+x+y)+2 w y) \end{aligned}$$ However, AM-GM gives us $$w x y(w+x+y) \leq \frac{1}{2}\left(\frac{x(w+x+y)+2 w y}{2}\right)^{2} \leq \frac{1}{8}$$ This proves $S(a, b, c, d) \geq-\frac{1}{8}$ for any $a, b, c, d \in \mathbb{R}$ such that $a>b>c>d$. Equality holds if and only if $w=y, x(w+x+y)=2 w y$ and $w x y(w+x+y)=\frac{1}{8}$. Solving these equations gives us $w^{4}=\frac{1}{16}$ which forces $w=\frac{1}{2}$ since $w>0$. Solving for $x$ gives us $x(x+1)=\frac{1}{2}$, and we will get $x=-\frac{1}{2}+\frac{\sqrt{3}}{2}$ as $x>0$. Plugging back gives us $d=-\frac{1}{4}-\frac{\sqrt{3}}{4}$, and this gives us $$(a, b, c, d)=\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}\right)$$ Thus, any cyclic permutation of the above solution will achieve the minimum equality. - Exactly 3 of $a-b, b-c, c-d, d-a$ are negative Since $a=\max \{a, b, c, d\}$, then $a-b$ has to be positive. So we must have $b<c<d<a$. Now, note that $$\begin{aligned} S(a, b, c, d) & =(a-b)(b-c)(c-d)(d-a) \\ & =(a-d)(d-c)(c-b)(b-a) \\ & =S(a, d, c, b) \end{aligned}$$ Now, note that $a>d>c>b$. By the previous case, $S(a, d, c, b) \geq-\frac{1}{8}$, which implies that $$S(a, b, c, d)=S(a, d, c, b) \geq-\frac{1}{8}$$ as well. Equality holds if and only if $$(a, b, c, d)=\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right)$$ and its cyclic permutation. |
4092 | 3 | 5 | 8 | During the regular season, Washington Redskins achieve a record of 10 wins and 6 losses. Compute the probability that their wins came in three streaks of consecutive wins, assuming that all possible arrangements of wins and losses are equally likely. (For example, the record LLWWWWWLWWLWWWLL contains three winning streaks, while WWWWWWWLLLLLLWWW has just two.) | \frac{315}{2002} | Suppse the winning streaks consist of $w_{1}, w_{2}$, and $w_{3}$ wins, in chronological order, where the first winning streak is preceded by $l_{0}$ consecutive losses and the $i$ winning streak is immediately succeeded by $l_{i}$ losses. Then $w_{1}, w_{2}, w_{3}, l_{1}, l_{2}>0$ are positive and $l_{0}, l_{3} \geq 0$ are nonnegative. The equations $$w_{1}+w_{2}+w_{3}=10 \quad \text { and } \quad\left(l_{0}+1\right)+l_{1}+l_{2}+\left(l_{3}+1\right)=8$$ are independent, and have $\binom{9}{2}$ and $\binom{7}{3}$ solutions, respectively. It follows that the answer is $$\frac{\binom{9}{2}\binom{7}{3}}{\binom{16}{6}}=\frac{315}{2002}$$ |
4095 | 4 | 4 | 8 | Bob's Rice ID number has six digits, each a number from 1 to 9, and any digit can be used any number of times. The ID number satisfies the following property: the first two digits is a number divisible by 2, the first three digits is a number divisible by 3, etc. so that the ID number itself is divisible by 6. One ID number that satisfies this condition is 123252. How many different possibilities are there for Bob's ID number? | 324 | We will count the number of possibilities for each digit in Bob's ID number, then multiply them to find the total number of possibilities for Bob's ID number. There are 3 possibilities for the first digit given any last 5 digits, because the entire number must be divisible by 3, so the sum of the digits must be divisible by 3. Because the first two digits are a number divisible by 2, the second digit must be \(2,4,6\), or 8, which is 4 possibilities. Because the first five digits are a number divisible by 5, the fifth digit must be a 5. Now, if the fourth digit is a 2, then the last digit has two choices, 2,8, and the third digit has 5 choices, \(1,3,5,7,9\). If the fourth digit is a 4, then the last digit must be a 6, and the third digit has 4 choices, \(2,4,6,8\). If the fourth digit is a 6, then the last digit must be a 4, and the third digit has 5 choices, \(1,3,5,7,9\). If the fourth digit is an 8, then the last digit has two choices, 2,8, and the third digit has 4 choices, \(2,4,6,8\). So there are a total of \(3 \cdot 4(2 \cdot 5+4+5+2 \cdot 4)=3 \cdot 4 \cdot 27=324\) possibilities for Bob's ID number. |
4096 | 3 | 5 | 8 | In how many distinct ways can you color each of the vertices of a tetrahedron either red, blue, or green such that no face has all three vertices the same color? (Two colorings are considered the same if one coloring can be rotated in three dimensions to obtain the other.) | 6 | If only two colors are used, there is only one possible arrangement up to rotation, so this gives 3 possibilities. If all three colors are used, then one is used twice. There are 3 ways to choose the color that is used twice. Say this color is red. Then the red vertices are on a common edge, and the green and blue vertices are on another edge. We see that either choice of arrangement of the green and blue vertices is the same up to rotation. Thus there are 6 possibilities total. |
4098 | 1 | 7 | 8 | The average of a set of distinct primes is 27. What is the largest prime that can be in this set? | 139 | Denote the set of these primes by $A$ and the number of elements in $A$ by n. There are 9 primes smaller than 27, namely $2,3,5,7,11,13,17,19$ and 23. Since 27 is odd and all primes except 2 are odd, $2 \notin A$. Thus the largest prime $p$ is at most $27 \cdot 9-3-5-7-11-13-17-19-23=145$, so $p \leq 141$. When the primes are $3,5,7,11,13,17,19,23,29,31$ and 139, their average is 27. Therefore $p=139$. |
4105 | 2 | 6 | 8 | In three-dimensional space, let $S$ be the region of points $(x, y, z)$ satisfying $-1 \leq z \leq 1$. Let $S_{1}, S_{2}, \ldots, S_{2022}$ be 2022 independent random rotations of $S$ about the origin ( $0,0,0$). The expected volume of the region $S_{1} \cap S_{2} \cap \cdots \cap S_{2022}$ can be expressed as $\frac{a \pi}{b}$, for relatively prime positive integers $a$ and $b$. Compute $100 a+b$. | 271619 | Consider a point $P$ of distance $r$ from the origin. The distance from the origin of a random projection of $P$ onto a line is uniform from 0 to $r$. Therefore, if $r<1$ then the probability of $P$ being in all the sets is 1, while for $r \geq 1$ it is $r^{-2022}$. Therefore the volume is $$\frac{4 \pi}{3}+4 \pi \int_{1}^{\infty} r^{2} r^{-2022} d r=4 \pi\left(\frac{1}{3}+\frac{1}{2019}\right)=\frac{2696 \pi}{2019}$$ |
4111 | 4 | 4 | 8 | Find the largest positive integer solution of the equation $\left\lfloor\frac{N}{3}\right\rfloor=\left\lfloor\frac{N}{5}\right\rfloor+\left\lfloor\frac{N}{7}\right\rfloor-\left\lfloor\frac{N}{35}\right\rfloor$. | 65 | For $N$ to be a solution, it is necessary that $\frac{N-2}{3}+\frac{N-34}{35} \leq \frac{N}{5}+\frac{N}{7}$, which simplifies to $N \leq 86$. However, if $N \geq 70$, then $N \leq 59$, contradicting $N \geq 70$. It follows that $N$ must be at most 69. Checking for $N \leq 69$, we find that when $N=65$, the equation holds. Thus the answer is $N=65$. |
4112 | 1 | 7 | 8 | $A B C D$ is a cyclic quadrilateral in which $A B=3, B C=5, C D=6$, and $A D=10 . M, I$, and $T$ are the feet of the perpendiculars from $D$ to lines $A B, A C$, and $B C$ respectively. Determine the value of $M I / I T$. | \frac{25}{9} | Quadrilaterals $A M I D$ and $D I C T$ are cyclic, having right angles $\angle A M D, \angle A I D$, and $\angle C I D, \angle C T D$ respectively. We see that $M, I$, and $T$ are collinear. For, $m \angle M I D=\pi-m \angle D A M=$ $\pi-m \angle D A B=m \angle B C D=\pi-m \angle D C T=\pi-m \angle D I T$. Therefore, Menelaus' theorem applied to triangle MTB and line $I C A$ gives $$\frac{M I}{I T} \cdot \frac{T C}{C B} \cdot \frac{B A}{A M}=1$$ On the other hand, triangle $A D M$ is similar to triangle $C D T$ since $\angle A M D \cong \angle C T D$ and $\angle D A M \cong$ $\angle D C T$ and thus $A M / C T=A D / C D$. It follows that $$\frac{M I}{I T}=\frac{B C \cdot A M}{A B \cdot C T}=\frac{B C \cdot A D}{A B \cdot C D}=\frac{5 \cdot 10}{3 \cdot 6}=\frac{25}{9}$$ Remarks. The line $M I T$, constructed in this problem by taking perpendiculars from a point on the circumcircle of $A B C$, is known as the Simson line. It is often helpful for us to use directed angles while angle chasing to avoid supplementary configuration issues, such as those arising while establishing the collinearity of $M, I$, and $T$. |
4123 | 3 | 5 | 8 | Bob the bomb-defuser has stumbled upon an active bomb. He opens it up, and finds the red and green wires conveniently located for him to cut. Being a seasoned member of the bomb-squad, Bob quickly determines that it is the green wire that he should cut, and puts his wirecutters on the green wire. But just before he starts to cut, the bomb starts to count down, ticking every second. Each time the bomb ticks, starting at time $t=15$ seconds, Bob panics and has a certain chance to move his wirecutters to the other wire. However, he is a rational man even when panicking, and has a $\frac{1}{2 t^{2}}$ chance of switching wires at time $t$, regardless of which wire he is about to cut. When the bomb ticks at $t=1$, Bob cuts whatever wire his wirecutters are on, without switching wires. What is the probability that Bob cuts the green wire? | \frac{23}{30} | Suppose Bob makes $n$ independent decisions, with probabilities of switching $p_{1}, p_{2}, \ldots, p_{n}$. Then in the expansion of the product $$P(x)=\left(p_{1}+\left(1-p_{1}\right) x\right)\left(p_{2}+\left(1-p_{2}\right) x\right) \cdots\left(p_{n}+\left(1-p_{n}\right) x\right)$$ the sum of the coefficients of even powers of $x$ gives the probability that Bob makes his original decision. This is just $(P(1)+P(-1)) / 2$, so the probability is just $$\frac{1+\left(1-\frac{1}{1515}\right)\left(1-\frac{1}{1414}\right) \cdots\left(1-\frac{1}{22}\right)}{2}=\frac{1+\frac{1416}{15151315} \cdots \frac{13}{22}}{2}=\frac{1+\frac{8}{15}}{2}=\frac{23}{30}$$ |
4128 | 4 | 4 | 8 | Circle $\Omega$ has radius 13. Circle $\omega$ has radius 14 and its center $P$ lies on the boundary of circle $\Omega$. Points $A$ and $B$ lie on $\Omega$ such that chord $A B$ has length 24 and is tangent to $\omega$ at point $T$. Find $A T \cdot B T$. | 56 | Let $M$ be the midpoint of chord $A B$; then $A M=B M=12$ and Pythagoras on triangle $A M O$ gives $M O=5$. Note that $\angle A O M=\angle A O B / 2=\angle A P B=\angle A P T+\angle T P B$ or $\tan (\angle A O M)=\tan (\angle A P T+\angle T P B)$. Applying the tangent addition formula, $\frac{A M}{M O} =\frac{\frac{A T}{T P}+\frac{B T}{T P}}{1-\frac{A T}{T P} \cdot \frac{B T}{T P}} =\frac{A B \cdot T P}{T P^{2}-A T \cdot B T}$ from which $A T \cdot B T=T P^{2}-A B \cdot T P \cdot M O / A M=14^{2}-24 \cdot 14 \cdot 5 / 12=56$. |
4129 | 1 | 7 | 8 | The Fibonacci numbers are defined recursively by $F_{0}=0, F_{1}=1$, and $F_{i}=F_{i-1}+F_{i-2}$ for $i \geq 2$. Given 15 wooden blocks of weights $F_{2}, F_{3}, \ldots, F_{16}$, compute the number of ways to paint each block either red or blue such that the total weight of the red blocks equals the total weight of the blue blocks. | 32 | Partition the blocks into sets $$\left\{F_{2}, F_{3}, F_{4}\right\},\left\{F_{5}, F_{6}, F_{7}\right\}, \ldots,\left\{F_{14}, F_{15}, F_{16}\right\}$$ We can show by bounding that $F_{16}$ belongs on the opposite side as $F_{15}$ and $F_{14}$, and, in general, that $F_{3 k+1}$ is on the opposite side as $F_{3 k}$ and $F_{3 k-1}$. Hence, it suffices to choose which side each of $F_{4}, F_{7}, \ldots, F_{16}$ go. This gives $2^{5}=32$ ways. |
4139 | 1 | 7 | 8 | Determine all pairs $(a, b)$ of integers with the property that the numbers $a^{2}+4 b$ and $b^{2}+4 a$ are both perfect squares. | (-4,-4),(-5,-6),(-6,-5),(0, k^{2}),(k^{2}, 0),(k, 1-k) | Without loss of generality, assume that $|b| \leq|a|$. If $b=0$, then $a$ must be a perfect square. So $(a=k^{2}, b=0)$ for each $k \in \mathbb{Z}$ is a solution. Now we consider the case $b \neq 0$. Because $a^{2}+4 b$ is a perfect square, the quadratic equation $x^{2}+a x-b=0$ has two non-zero integral roots $x_{1}, x_{2}$. Then $x_{1}+x_{2}=-a$ and $x_{1} x_{2}=-b$, and from this it follows that $\frac{1}{\left|x_{1}\right|}+\frac{1}{\left|x_{2}\right|} \geq\left|\frac{1}{x_{1}}+\frac{1}{x_{2}}\right|=\frac{|a|}{|b|} \geq 1$. Hence there is at least one root, say $x_{1}$, such that $\left|x_{1}\right| \leq 2$. There are the following possibilities. (1) $x_{1}=2$. Substituting $x_{1}=2$ into the quadratic equation we get $b=2a+4$. So we have $b^{2}+4a=(2a+4)^{2}+4a=4a^{2}+20a+16=(2a+5)^{2}-9$. It is easy to see that the solution in non-negative integers of the equation $x^{2}-9=y^{2}$ is $(3,0)$. Hence $2a+5= \pm 3$. From this we obtain $a=-4, b=-4$ and $a=-1, b=2$. The latter should be rejected because of the assumption $|a| \geq|b|$. (2) $x_{1}=-2$. Substituting $x_{1}=-2$ into the quadratic equation we get $b=4-2a$. Hence $b^{2}+4a=4a^{2}-12a+16=(2a-3)^{2}+7$. It is easy to show that the solution in non-negative integers of the equation $x^{2}+7=y^{2}$ is $(3,4)$. Hence $2a-3= \pm 3$. From this we obtain $a=3, b=-2$. (3) $x_{1}=1$. Substituting $x_{1}=1$ into the quadratic equation we get $b=a+1$. Hence $b^{2}+4a=a^{2}+6a+1=(a+3)^{2}-8$. It is easy to show that the solution in non-negative integers of the equation $x^{2}-8=y^{2}$ is $(3,1)$. Hence $a+3= \pm 3$. From this we obtain $a=-6, b=-5$. (4) $x_{1}=-1$. Substituting $x_{1}=-1$ into the quadratic equation we get $b=1-a$. Then $a^{2}+4b=(a-2)^{2}, b^{2}+4a=(a+1)^{2}$. Consequently, $a=k, b=1-k(k \in \mathbb{Z})$ is a solution. Testing these solutions and by symmetry we obtain the following solutions $(-4,-4),(-5,-6),(-6,-5),\left(0, k^{2}\right),\left(k^{2}, 0\right),(k, 1-k)$ where $k$ is an arbitrary integer. |
4142 | 3 | 5 | 8 | In a game, \(N\) people are in a room. Each of them simultaneously writes down an integer between 0 and 100 inclusive. A person wins the game if their number is exactly two-thirds of the average of all the numbers written down. There can be multiple winners or no winners in this game. Let \(m\) be the maximum possible number such that it is possible to win the game by writing down \(m\). Find the smallest possible value of \(N\) for which it is possible to win the game by writing down \(m\) in a room of \(N\) people. | 34 | Since the average of the numbers is at most 100, the winning number is an integer which is at most two-thirds of 100, or at most 66. This is achieved in a room with 34 people, in which 33 people pick 100 and one person picks 66, so the average number is 99. Furthermore, this cannot happen with less than 34 people. If the winning number is 66 and there are \(N\) people, the sum of the numbers must be 99 then we must have that \(99 N \leq 66+100(N-1)\), which reduces to \(N \geq 34\). |
4143 | 2 | 6 | 8 | Descartes's Blackjack: How many integer lattice points (points of the form $(m, n)$ for integers $m$ and $n$) lie inside or on the boundary of the disk of radius 2009 centered at the origin? | 12679605 | The number of lattice points inside or on the boundary of a circle with radius $r$ centered at the origin can be approximated using the formula $\pi r^2 + \text{error term}$. For a circle with radius 2009, this results in approximately 12679605 lattice points. |
4152 | 3 | 5 | 8 | On a board the following six vectors are written: \((1,0,0), \quad(-1,0,0), \quad(0,1,0), \quad(0,-1,0), \quad(0,0,1), \quad(0,0,-1)\). Given two vectors \(v\) and \(w\) on the board, a move consists of erasing \(v\) and \(w\) and replacing them with \(\frac{1}{\sqrt{2}}(v+w)\) and \(\frac{1}{\sqrt{2}}(v-w)\). After some number of moves, the sum of the six vectors on the board is \(u\). Find, with proof, the maximum possible length of \(u\). | 2 \sqrt{3} | For a construction, note that one can change \((1,0,0),(-1,0,0) \rightarrow(\sqrt{2}, 0,0),(0,0,0) \rightarrow(1,0,0),(1,0,0)\) and similarly for \((0,1,0),(0,-1,0)\) and \((0,0,1),(0,0,-1)\). Then \(u=(2,2,2)\). For the bound, argue as follows: let the vectors be \(v_{1}, \ldots, v_{6}, n=(x, y, z)\) be any unit vector, and \(S=\sum_{i}\left(n \cdot v_{i}\right)^{2}\), where the sum is over all vectors on the board. We claim that \(S\) is invariant. Indeed, we have \(\left(n \cdot \frac{1}{\sqrt{2}}(v+w)\right)^{2}+\left(n \cdot \frac{1}{\sqrt{2}}(v-w)\right)^{2} =\left(\frac{n \cdot v+n \cdot w}{\sqrt{2}}\right)^{2}+\left(\frac{n \cdot v-n \cdot w}{\sqrt{2}}\right)^{2} =\frac{2(n \cdot v)^{2}+2(n \cdot w)^{2}}{2} =(n \cdot v)^{2}+(n \cdot w)^{2}\). Also, at the beginning we have \(S=2x^{2}+2y^{2}+2z^{2}=2\). Therefore we must always have \(S=2\). Thus, by the Cauchy-Schwarz inequality we have \(n \cdot u=\sum n \cdot v_{i} \leq \sqrt{\sum_{i}\left(n \cdot v_{i}\right)^{2}} \sqrt{6}=\sqrt{12}=2\sqrt{3}\). But since \(n\) is arbitrary, this implies that \(|u| \leq 2\sqrt{3}\); otherwise we could pick \(n=u /|u|\) and reach a contradiction. |
4157 | 4 | 4 | 8 | Find the value of \(\sum_{k=1}^{60} \sum_{n=1}^{k} \frac{n^{2}}{61-2 n}\). | -18910 | Change the order of summation and simplify the inner sum: \(\sum_{k=1}^{60} \sum_{n=1}^{k} \frac{n^{2}}{61-2 n} =\sum_{n=1}^{60} \sum_{k=n}^{60} \frac{n^{2}}{61-2 n} =\sum_{n=1}^{60} \frac{n^{2}(61-n)}{61-2 n}\). Then, we rearrange the sum to add the terms corresponding to \(n\) and \(61-n\): \(\sum_{n=1}^{60} \frac{n^{2}(61-n)}{61-2 n} =\sum_{n=1}^{30}\left(\frac{n^{2}(61-n)}{61-2 n}+\frac{(61-n)^{2}(61-(61-n))}{61-2(61-n)}\right) =\sum_{n=1}^{30} \frac{n^{2}(61-n)-n(61-n)^{2}}{61-2 n} =\sum_{n=1}^{30} \frac{n(61-n)(n-(61-n))}{61-2 n} =\sum_{n=1}^{30}-n(61-n) =\sum_{n=1}^{30} n^{2}-61 n\). Finally, using the formulas for the sum of the first \(k\) squares and sum of the first \(k\) positive integers, we conclude that this last sum is \(\frac{30(31)(61)}{6}-61 \frac{30(31)}{2}=-18910\). So, the original sum evaluates to -18910. |
4166 | 4 | 4 | 8 | The sequence $\left\{a_{n}\right\}_{n \geq 1}$ is defined by $a_{n+2}=7 a_{n+1}-a_{n}$ for positive integers $n$ with initial values $a_{1}=1$ and $a_{2}=8$. Another sequence, $\left\{b_{n}\right\}$, is defined by the rule $b_{n+2}=3 b_{n+1}-b_{n}$ for positive integers $n$ together with the values $b_{1}=1$ and $b_{2}=2$. Find \operatorname{gcd}\left(a_{5000}, b_{501}\right). | 89 | We show by induction that $a_{n}=F_{4 n-2}$ and $b_{n}=F_{2 n-1}$, where $F_{k}$ is the $k$ th Fibonacci number. The base cases are clear. As for the inductive steps, note that $$F_{k+2}=F_{k+1}+F_{k}=2 F_{k}+F_{k-1}=3 F_{k}-F_{k-2}$$ and $$F_{k+4}=3 F_{k+2}-F_{k}=8 F_{k}+3 F_{k-2}=7 F_{k}-F_{k-4}$$ We wish to compute the greatest common denominator of $F_{19998}$ and $F_{1001}$. The Fibonacci numbers satisfy the property that \operatorname{gcd}\left(F_{m}, F_{n}\right)=F_{\operatorname{gcd}(m, n)}$, which can be proven by noting that they are periodic modulo any positive integer. So since \operatorname{gcd}(19998,1001)=11, the answer is $F_{11}=89$. |
4183 | 4 | 4 | 8 | Let $\mathbb{N}$ be the set of positive integers, and let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying $f(1)=1$ and for $n \in \mathbb{N}, f(2 n)=2 f(n)$ and $f(2 n+1)=2 f(n)-1$. Determine the sum of all positive integer solutions to $f(x)=19$ that do not exceed 2019. | 1889 | For $n=2^{a_{0}}+2^{a_{1}}+\cdots+2^{a_{k}}$ where $a_{0}>a_{1}>\cdots>a_{k}$, we can show that $f(n)=2^{a_{0}}-2^{a_{1}}-\cdots-2^{a_{k}}=2^{a_{0}+1}-n$ by induction: the base case $f(1)=1$ clearly holds; for the inductive step, when $n$ is even we note that $f(n)=2 f\left(\frac{n}{2}\right)=2\left(2^{a_{0}}-\frac{n}{2}\right)=2^{a_{0}+1}-n$ as desired, and when $n$ is odd we also have $f(n)=2 f\left(\frac{n-1}{2}\right)-1=2\left(2^{a_{0}}-\frac{n-1}{2}\right)-1=2^{a_{0}+1}-n$, again as desired. Since $19=f(n) \leq 2^{a_{0}} \leq n$, we have $a_{0} \geq 5$ and $n=2^{a_{0}+1}-19 \leq 2019$ gives $a_{0} \leq 9$. So the answer is $\sum_{a=5}^{9}\left(2^{a+1}-19\right)=\left(2^{11}-2^{6}\right)-19 \cdot 5=1889$. |
4185 | 4 | 4 | 8 | A sequence consists of the digits $122333444455555 \ldots$ such that each positive integer $n$ is repeated $n$ times, in increasing order. Find the sum of the 4501st and 4052nd digits of this sequence. | 13 | Note that $n$ contributes $n \cdot d(n)$ digits, where $d(n)$ is the number of digits of $n$. Then because $1+\cdots+99=4950$, we know that the digits of interest appear amongst copies of two digit numbers. Now for $10 \leq n \leq 99$, the number of digits in the subsequence up to the last copy of $n$ is $$1+2+3+\cdots+9+2 \cdot(10+\cdots+n)=2 \cdot(1+\cdots+n)-45=n^{2}+n-45$$ Since $67^{2}+67-45=4511$, the two digits are 6 and 7 in some order, so have sum 13. |
4191 | 4 | 4 | 8 | Let $S$ be the set of integers of the form $2^{x}+2^{y}+2^{z}$, where $x, y, z$ are pairwise distinct non-negative integers. Determine the 100th smallest element of $S$. | 577 | S is the set of positive integers with exactly three ones in its binary representation. The number of such integers with at most $d$ total bits is \binom{d}{3}$, and noting that \binom{9}{3}=84$ and \binom{10}{3}=120$, we want the 16th smallest integer of the form $2^{9}+2^{x}+2^{y}$, where $y<x<9$. Ignoring the $2^{9}$ term, there are \binom{d^{\prime}}{2}$ positive integers of the form $2^{x}+2^{y}$ with at most $d^{\prime}$ total bits. Because \binom{6}{2}=15$, our answer is $2^{9}+2^{6}+2^{0}=577$. (By a bit, we mean a digit in base 2 .) |
4195 | 1 | 7 | 8 | Rachel has the number 1000 in her hands. When she puts the number $x$ in her left pocket, the number changes to $x+1$. When she puts the number $x$ in her right pocket, the number changes to $x^{-1}$. Each minute, she flips a fair coin. If it lands heads, she puts the number into her left pocket, and if it lands tails, she puts it into her right pocket. She then takes the new number out of her pocket. If the expected value of the number in Rachel's hands after eight minutes is $E$, then compute $\left\lfloor\frac{E}{10}\right\rfloor$. | 13 | Call a real number very large if $x \in[1000,1008]$, very small if $x \in\left[0, \frac{1}{1000}\right]$, and medium-sized if $x \in\left[\frac{1}{8}, 8\right]$. Every number Rachel is ever holding after at most 8 steps will fall under one of these categories. Therefore the main contribution to $E$ will come from the probability that Rachel is holding a number at least 1000 at the end. Note that if her number ever becomes medium-sized, it will never become very large or very small again. Therefore the only way her number ends up above 1000 is if the sequence of moves consists of $x \rightarrow x+1$ moves and consecutive pairs of $x \rightarrow x^{-1}$ moves. Out of the 256 possible move sequences, the number of ways for the number to stay above 1000 is the number of ways of partitioning 8 into an ordered sum of 1 and 2, or the ninth Fibonacci number $F_{9}=34$. Therefore $$\frac{34}{256} \cdot 1000 \leq E \leq \frac{34}{256} \cdot 1000+8$$ where $\frac{34}{256} \cdot 1000 \approx 132.8$. Furthermore, the extra contribution will certainly not exceed 7 , so we get that $\left\lfloor\frac{E}{10}\right\rfloor=13$. |
4207 | 2 | 6 | 8 | Acute triangle $A B C$ has circumcenter $O$. The bisector of $\angle A B C$ and the altitude from $C$ to side $A B$ intersect at $X$. Suppose that there is a circle passing through $B, O, X$, and $C$. If $\angle B A C=n^{\circ}$, where $n$ is a positive integer, compute the largest possible value of $n$. | 67 | We have $\angle X B C=B / 2$ and $\angle X C B=90^{\circ}-B$. Thus, $\angle B X C=90^{\circ}+B / 2$. We have $\angle B O C=2 A$, so $$90^{\circ}+B / 2=2 A$$ This gives $B=4 A-180^{\circ}$, which gives $C=360^{\circ}-5 A$. In order for $0^{\circ}<B<90^{\circ}$, we need $45^{\circ}<A<67.5^{\circ}$. In order for $0^{\circ}<C<90^{\circ}$, we require $54^{\circ}<A<72^{\circ}$. The largest integer value in degrees satisfying these inequalities is $A=67^{\circ}$. |
4212 | 2 | 6 | 8 | Let $z$ be a non-real complex number with $z^{23}=1$. Compute $$ \sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}} $$ | 46 / 3 | First solution: Note that $$ \sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}}=\frac{1}{3}+\sum_{k=1}^{22} \frac{1-z^{k}}{1-z^{3 k}}=\frac{1}{3}+\sum_{k=1}^{22} \frac{1-\left(z^{24}\right)^{k}}{1-z^{3 k}}=\frac{1}{3}+\sum_{k=1}^{22} \sum_{\ell=0}^{7} z^{3 k \ell} $$ 3 and 23 are prime, so every non-zero residue modulo 23 appears in an exponent in the last sum exactly 7 times, and the summand 1 appears 22 times. Because the sum of the 23 rd roots of unity is zero, our answer is $\frac{1}{3}+(22-7)=\frac{46}{3}$. Second solution: For an alternate approach, we first prove the following identity for an arbitrary complex number $a$ : $$ \sum_{k=0}^{22} \frac{1}{a-z^{k}}=\frac{23 a^{22}}{a^{23}-1} $$ To see this, let $f(x)=x^{23}-1=(x-1)(x-z)\left(x-z^{2}\right) \ldots\left(x-z^{22}\right)$. Note that the sum in question is merely $\frac{f^{\prime}(a)}{f(a)}$, from which the identity follows. Now, returning to our original sum, let $\omega \neq 1$ satisfy $\omega^{3}=1$. Then $$ \begin{aligned} \sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}} & =\frac{1}{\omega^{2}-\omega} \sum_{k=0}^{22} \frac{1}{\omega-z^{k}}-\frac{1}{\omega^{2}-z^{k}} \ & =\frac{1}{\omega^{2}-\omega}\left(\sum_{k=0}^{22} \frac{1}{\omega-z^{k}}-\sum_{k=0}^{22} \frac{1}{\omega^{2}-z^{k}}\right) \ & =\frac{1}{\omega^{2}-\omega}\left(\frac{23 \omega^{22}}{\omega^{23}-1}-\frac{23 \omega^{44}}{\omega^{46}-1}\right) \ & =\frac{23}{\omega^{2}-\omega}\left(\frac{\omega}{\omega^{2}-1}-\frac{\omega^{2}}{\omega-1}\right) \ & =\frac{23}{\omega^{2}-\omega} \frac{\left(\omega^{2}-\omega\right)-\left(\omega-\omega^{2}\right)}{2-\omega-\omega^{2}} \ & =\frac{46}{3} \end{aligned} $$ |
4215 | 2 | 6 | 8 | Suppose $a, b, c$, and $d$ are pairwise distinct positive perfect squares such that $a^{b}=c^{d}$. Compute the smallest possible value of $a+b+c+d$. | 305 | Note that if $a$ and $c$ are divisible by more than one distinct prime, then we can just take the prime powers of a specific prime. Thus, assume $a$ and $c$ are powers of a prime $p$. Assume $a=4^{x}$ and $c=4^{y}$. Then $x b=y d$. Because $b$ and $d$ are squares, the ratio of $x$ to $y$ is a square, so assume $x=1$ and $y=4$. We can't take $b=4$ and $c=1$, but we instead can take $b=36$ and $c=9$. It can be checked that other values of $x$ and $y$ are too big. This gives $4^{36}=256^{9}$, which gives a sum of 305. If $a$ and $c$ are powers of 9 , then $\max (a, c) \geq 9^{4}$, which is already too big. Thus, 305 is optimal. |
4216 | 2 | 6 | 8 | Fran writes the numbers \(1,2,3, \ldots, 20\) on a chalkboard. Then she erases all the numbers by making a series of moves; in each move, she chooses a number \(n\) uniformly at random from the set of all numbers still on the chalkboard, and then erases all of the divisors of \(n\) that are still on the chalkboard (including \(n\) itself). What is the expected number of moves that Fran must make to erase all the numbers? | \frac{131}{10} | For each \(n, 1 \leq n \leq 20\), consider the first time that Fran chooses one of the multiples of \(n\). It is in this move that \(n\) is erased, and all the multiples of \(n\) at most 20 are equally likely to be chosen for this move. Hence this is the only move in which Fran could possibly choose \(n\); since there are \(\lfloor 20 / n\rfloor\) multiples of \(n\) at most 20, this means that the probability that \(n\) is ever chosen is \(1 /\lfloor 20 / n\rfloor\). Therefore the expected number of moves is \(E =\sum_{n=1}^{20} \frac{1}{\lfloor 20 / n\rfloor} =\frac{1}{20}+\frac{1}{10}+\frac{1}{6}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+4\left(\frac{1}{2}\right)+10(1)=\frac{131}{10}\). (This sum is easier to compute than it may seem, if one notes that \(1 / 20+1 / 5+1 / 4=1 / 2\) and \(1 / 6+1 / 3=1 / 2)\) |
4225 | 1 | 7 | 8 | Determine the number of ways to select a sequence of 8 sets $A_{1}, A_{2}, \ldots, A_{8}$, such that each is a subset (possibly empty) of \{1,2\}, and $A_{m}$ contains $A_{n}$ if $m$ divides $n$. | 2025 | Consider an arbitrary $x \in\{1,2\}$, and let us consider the number of ways for $x$ to be in some of the sets so that the constraints are satisfied. We divide into a few cases: - Case: $x \notin A_{1}$. Then $x$ cannot be in any of the sets. So there is one possibility. - Case: $x \in A_{1}$ but $x \notin A_{2}$. Then the only other sets that $x$ could be in are $A_{3}, A_{5}, A_{7}$, and $x$ could be in some collection of them. There are 8 possibilities in this case. - Case: $x \in A_{2}$. Then $x \in A_{1}$ automatically. There are 4 independent choices to be make here: (1) whether $x \in A_{5} ;(2)$ whether $x \in A_{7} ;(3)$ whether $x \in A_{3}$, and if yes, whether $x \in A_{6}$; (4) whether $x \in A_{4}$, and if yes, whether $x \in A_{8}$. There are $2 \times 2 \times 3 \times 3=36$ choices here. Therefore, there are $1+8+36=45$ ways to place $x$ into some of the sets. Since the choices for $x=1$ and $x=2$ are made independently, we see that the total number of possibilities is $45^{2}=2025$. |
4226 | 4 | 4 | 8 | Let $P_{1}, P_{2}, \ldots, P_{8}$ be 8 distinct points on a circle. Determine the number of possible configurations made by drawing a set of line segments connecting pairs of these 8 points, such that: (1) each $P_{i}$ is the endpoint of at most one segment and (2) two no segments intersect. (The configuration with no edges drawn is allowed.) | 323 | Let $f(n)$ denote the number of valid configurations when there are $n$ points on the circle. Let $P$ be one of the points. If $P$ is not the end point of an edge, then there are $f(n-1)$ ways to connect the remaining $n-1$ points. If $P$ belongs to an edge that separates the circle so that there are $k$ points on one side and $n-k-2$ points on the other side, then there are $f(k) f(n-k-2)$ ways of finishing the configuration. Thus, $f(n)$ satisfies the recurrence relation $$f(n)=f(n-1)+f(0) f(n-2)+f(1) f(n-3)+f(2) f(n-4)+\cdots+f(n-2) f(0), n \geq 2$$ The initial conditions are $f(0)=f(1)=1$. Using the recursion, we find that $f(2)=2, f(3)=4, f(4)=$ $9, f(5)=21, f(6)=51, f(7)=127, f(8)=323$. |
4230 | 4 | 4 | 8 | Compute the sum of all positive integers $n$ such that $50 \leq n \leq 100$ and $2 n+3$ does not divide $2^{n!}-1$. | 222 | We claim that if $n \geq 10$, then $2 n+3 \nmid 2^{n!}-1$ if and only if both $n+1$ and $2 n+3$ are prime. If both $n+1$ and $2 n+3$ are prime, then assume $2 n+3 \mid 2^{n!}-1$. By Fermat Little Theorem, $2 n+3 \mid 2^{2 n+2}+1$. However, since $n+1$ is prime, $\operatorname{gcd}(2 n+2, n!)=2$, so $2 n+3 \mid 2^{2}-1=3$, a contradiction. If $2 n+3$ is composite, then $\varphi(2 n+3)$ is even and is at most $2 n$, so $\varphi(2 n+3) \mid n$ !, done. If $n+1$ is composite but $2 n+3$ is prime, then $2 n+2 \mid n$ !, so $2 n+3 \mid 2^{n!}-1$. The prime numbers between 50 and 100 are $53,59,61,67,71,73,79,83,89,97$. If one of these is $n+1$, then the only numbers that make $2 n+3$ prime are 53,83 , and 89 , making $n$ one of 52,82 , and 88 . These sum to 222. |
4232 | 1 | 7 | 8 | Determine the number of 8-tuples of nonnegative integers $\left(a_{1}, a_{2}, a_{3}, a_{4}, b_{1}, b_{2}, b_{3}, b_{4}\right)$ satisfying $0 \leq a_{k} \leq k$, for each $k=1,2,3,4$, and $a_{1}+a_{2}+a_{3}+a_{4}+2 b_{1}+3 b_{2}+4 b_{3}+5 b_{4}=19$. | 1540 | For each $k=1,2,3,4$, note that set of pairs $\left(a_{k}, b_{k}\right)$ with $0 \leq a_{k} \leq k$ maps bijectively to the set of nonnegative integers through the map $\left(a_{k}, b_{k}\right) \mapsto a_{k}+(k+1) b_{k}$, as $a_{k}$ is simply the remainder of $a_{k}+(k+1) b_{k}$ upon division by $k+1$. By letting $x_{k}=a_{k}+(k+1) b_{k}$, we see that the problem is equivalent to finding the number of quadruples of nonnegative integers $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ such that $x_{1}+x_{2}+x_{3}+x_{4}=19$. This is the same as finding the number of quadruples of positive integers $\left(x_{1}+1, x_{2}+1, x_{3}+1, x_{4}+1\right)$ such that $x_{1}+x_{2}+x_{3}+x_{4}=23$. By a standard "dots and bars" argument, we see that the answer is $\binom{22}{3}=1540$. |
4233 | 1 | 7 | 8 | Michel starts with the string $H M M T$. An operation consists of either replacing an occurrence of $H$ with $H M$, replacing an occurrence of $M M$ with $M O M$, or replacing an occurrence of $T$ with $M T$. For example, the two strings that can be reached after one operation are $H M M M T$ and $H M O M T$. Compute the number of distinct strings Michel can obtain after exactly 10 operations. | 144 | Each final string is of the form $H M x M T$, where $x$ is a string of length 10 consisting of $M \mathrm{~s}$ and $O$ s. Further, no two $O \mathrm{~s}$ can be adjacent. It is not hard to prove that this is a necessary and sufficient condition for being a final string. Let $f(n)$ be the number of strings of length $n$ consisting of $M$ s and $O$ where no two $O$ s are adjacent. Any such string of length $n+2$ must either end in $M$, in which case removing the $M$ results in a valid string of length $n+1$, or $M O$, in which case removing the $M O$ results in a valid string of length $n$. Therefore, $f(n+2)=f(n)+f(n+1)$. Since $f(1)=2$ and $f(2)=3$, applying the recursion leads to $f(10)=144$. |
4238 | 3 | 5 | 8 | A bug is on a corner of a cube. A healthy path for the bug is a path along the edges of the cube that starts and ends where the bug is located, uses no edge multiple times, and uses at most two of the edges adjacent to any particular face. Find the number of healthy paths. | 6 | There are 6 symmetric ways to choose the first two edges on the path. After these are chosen, all subsequent edges are determined, until the starting corner is reached once again. |
4239 | 1 | 7 | 8 | A polynomial $P$ of degree 2015 satisfies the equation $P(n)=\frac{1}{n^{2}}$ for $n=1,2, \ldots, 2016$. Find \lfloor 2017 P(2017)\rfloor. | -9 | Let $Q(x)=x^{2} P(x)-1$. Then $Q(n)=n^{2} P(n)-1=0$ for $n=1,2, \ldots, 2016$, and $Q$ has degree 2017 . Thus we may write $$Q(x)=x^{2} P(x)-1=(x-1)(x-2) \ldots(x-2016) L(x)$$ where $L(x)$ is some linear polynomial. Then $Q(0)=-1=(-1)(-2) \ldots(-2016) L(0)$, so $L(0)=-\frac{1}{2016!}$. Now note that $$\begin{aligned} Q^{\prime}(x) & =x^{2} P^{\prime}(x)+2 x P(x) \\ & =\sum_{i=1}^{2016}(x-1) \ldots(x-(i-1))(x-(i+1)) \ldots(x-2016) L(x)+(x-1)(x-2) \ldots(x-2016) L^{\prime}(x) \end{aligned}$$ Thus $$Q^{\prime}(0)=0=L(0)\left(\frac{2016!}{-1}+\frac{2016!}{-2}+\ldots+\frac{2016!}{-2016}\right)+2016!L^{\prime}(0)$$ whence $L^{\prime}(0)=L(0)\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{2016}\right)=-\frac{H_{2016}}{2016!}$, where $H_{n}$ denotes the $n$th harmonic number. As a result, we have $L(x)=-\frac{H_{2016} x+1}{2016!}$. Then $$Q(2017)=2017^{2} P(2017)-1=2016!\left(-\frac{2017 H_{2016}+1}{2016!}\right)$$ which is $-2017 H_{2016}-1$. Thus $$P(2017)=\frac{-H_{2016}}{2017}$$ From which we get $2017 P(2017)=-H_{2016}$. It remains to approximate $H_{2016}$. We alter the well known approximation $$H_{n} \approx \int_{1}^{n} \frac{1}{x} d x=\log x$$ to $$H_{n} \approx 1+\frac{1}{2}+\int_{3}^{n} \frac{1}{x} d x=1+\frac{1}{2}+\log (2016)-\log (3) \approx \log (2016)+\frac{1}{2}$$ so that it suffices to lower bound $\log (2016)$. Note that $e^{3} \approx 20$, which is close enough for our purposes. Then $e^{6} \approx 400 \Longrightarrow e^{7} \approx 1080$, and $e^{3} \approx 20<2^{5} \Longrightarrow e^{0.6}<<2 \Longrightarrow e^{7.6}<2016$, so that $\log (2016)>7.6$. It follows that $H_{2016} \approx \log (2016)+0.5=7.6+0.5>8$ (of course these are loose estimates, but more than good enough for our purposes). Thus $-9<2017 P(2017)<-8$, making our answer -9 . Alternatively, a well-read contestant might know that $H_{n} \approx \log n+\gamma$, where $\gamma \approx .577$ is the EulerMascheroni constant. The above solution essentially approximates $\gamma$ as 0.5 which is good enough for our purposes. |
4243 | 4 | 4 | 8 | Kermit the frog enjoys hopping around the infinite square grid in his backyard. It takes him 1 Joule of energy to hop one step north or one step south, and 1 Joule of energy to hop one step east or one step west. He wakes up one morning on the grid with 100 Joules of energy, and hops till he falls asleep with 0 energy. How many different places could he have gone to sleep? | 10201 | It is easy to see that the coordinates of the frog's final position must have the same parity. Suppose that the frog went to sleep at $(x, y)$. Then, we have that $-100 \leq y \leq 100$ and $|x| \leq 100-|y|$, so $x$ can take on the values $-100+|y|,-98+|y|, \ldots, 100-|y|$. There are $101-|y|$ such values, so the total number of such locations is $$\sum_{y=-100}^{100} 101-|y|=201 \cdot 101-2 \cdot \frac{100(100+1)}{2}=101^{2}=10201$$ |
4254 | 2 | 6 | 8 | Richard starts with the string HHMMMMTT. A move consists of replacing an instance of HM with MH , replacing an instance of MT with TM, or replacing an instance of TH with HT. Compute the number of possible strings he can end up with after performing zero or more moves. | 70 | The key claim is that the positions of the Ms fully determines the end configuration. Indeed, since all Hs are initially left of all Ts, the only successful swaps that can occur will involve Ms. So, picking $\binom{8}{4}=70$ spots for Ms and then filling in the remaining 4 spots with Hs first and then Ts gives all possible arrangements. It is not hard to show that all of these arrangements are also achievable; just greedily move Ms to their target positions. |
4255 | 1 | 7 | 8 | Five cards labeled $1,3,5,7,9$ are laid in a row in that order, forming the five-digit number 13579 when read from left to right. A swap consists of picking two distinct cards, and then swapping them. After three swaps, the cards form a new five-digit number $n$ when read from left to right. Compute the expected value of $n$. | 50308 | For a given card, let $p(n)$ denote the probability that it is in its original position after $n$ swaps. Then $p(n+1)=p(n) \cdot \frac{3}{5}+(1-p(n)) \cdot \frac{1}{10}$, by casework on whether the card is in the correct position or not after $n$ swaps. In particular, $p(0)=1, p(1)=3 / 5, p(2)=2 / 5$, and $p(3)=3 / 10$. For a certain digit originally occupied with the card labeled $d$, we see that, at the end of the process, the card at the digit is $d$ with probability $3 / 10$ and equally likely to be one of the four non- $d$ cards with probability $7 / 10$. Thus the expected value of the card at this digit is $$\frac{3 d}{10}+\frac{7}{10} \frac{25-d}{4}=\frac{12 d+175-7 d}{40}=\frac{d+35}{8}$$ By linearity of expectation, our final answer is therefore $$\frac{13579+35 \cdot 11111}{8}=\frac{402464}{8}=50308$$ |
4259 | 2 | 6 | 8 | Let \(A B C\) be a triangle with \(\angle A=18^{\circ}, \angle B=36^{\circ}\). Let \(M\) be the midpoint of \(A B, D\) a point on ray \(C M\) such that \(A B=A D ; E\) a point on ray \(B C\) such that \(A B=B E\), and \(F\) a point on ray \(A C\) such that \(A B=A F\). Find \(\angle F D E\). | 27 | Let \(\angle A B D=\angle A D B=x\), and \(\angle D A B=180-2 x\). In triangle \(A C D\), by the law of sines, \(C D=\frac{A D}{\sin \angle A C M} \cdot \sin 198-2 x\), and by the law of sines in triangle \(B C D, C D=\frac{B D}{\sin \angle B C M} \cdot \sin x+36\). Combining the two, we have \(2 \cos x=\frac{B D}{A D}=\frac{\sin 198-2 x}{\sin x+36} \cdot \frac{\sin \angle B C M}{\sin \angle A C M}\). But by the ratio lemma, \(1=\frac{M B}{M A}=\frac{C B}{C A} \frac{\sin \angle B C M}{\sin \angle A C M}\), meaning that \(\frac{\sin \angle B C M}{\sin \angle A C M}=\frac{C A}{C B}=\frac{\sin 36}{\sin 18}=2 \cos 18\). Plugging this in and simplifying, we have \(2 \cos x=\frac{\sin 198-2 x}{\sin x+36} \cdot 2 \cos 18=\frac{\cos 108-2 x}{\cos 54-x} \cdot 2 \cos 18\), so that \(\frac{\cos x}{\cos 18}=\frac{\cos 108-2 x}{\cos 54-x}\). We see that \(x=36^{\circ}\) is a solution to this equation, and by carefully making rough sketches of both functions, we can convince ourselves that this is the only solution where \(x\) is between 0 and 90 degrees. Therefore \(\angle A B D=\angle A D B=36\), \(\angle D A B=108\). Simple angle chasing yields \(\angle A E B=72, \angle E C A=54, \angle E A C=54, \angle E A B=72\), making \(D, A\), and \(E\) collinear, and so \(\angle B D E=36\). And because \(A F=A B=A D, \angle F D B=1 / 2 \angle F A B=9\), so \(\angle F D E=36-9=27\). |
4262 | 1 | 7 | 8 | Let $A, E, H, L, T$, and $V$ be chosen independently and at random from the set $\left\{0, \frac{1}{2}, 1\right\}$. Compute the probability that $\lfloor T \cdot H \cdot E\rfloor=L \cdot A \cdot V \cdot A$. | \frac{55}{81} | There are $3^{3}-2^{3}=19$ ways to choose $L, A$, and $V$ such that $L \cdot A \cdot V \cdot A=0$, since at least one of $\{L, A, V\}$ must be 0 , and $3^{3}-1=26$ ways to choose $T, H$, and $E$ such that $\lfloor T \cdot H \cdot E\rfloor=0$, since at least one of $\{T, H, E\}$ must not be 1 , for a total of $19 \cdot 26=494$ ways. There is only one way to make $\lfloor T \cdot H \cdot E\rfloor=L \cdot A \cdot V \cdot A=1$, namely setting every variable equal to 1 , so there are 495 total ways that work out of a possible $3^{6}=729$, for a probability of $\frac{55}{81}$. |
4264 | 1 | 7 | 8 | You are trapped in ancient Japan, and a giant enemy crab is approaching! You must defeat it by cutting off its two claws and six legs and attacking its weak point for massive damage. You cannot cut off any of its claws until you cut off at least three of its legs, and you cannot attack its weak point until you have cut off all of its claws and legs. In how many ways can you defeat the giant enemy crab? | 14400 | The answer is given by $6!2!\binom{5}{2}$, because we can cut off the claws and legs in any order and there are $\binom{5}{2}$ ways to decide when to cut off the two claws (since we can do it at any time among the last 5 cuts). |
4265 | 3 | 5 | 8 | Compute the number of ways to color 3 cells in a $3 \times 3$ grid so that no two colored cells share an edge. | 22 | If the middle square is colored, then two of the four corner squares must be colored, and there are $\binom{4}{2}=6$ ways to do this. If the middle square is not colored, then after coloring one of the 8 other squares, there are always 6 ways to place the other two squares. However, the number of possibilities is overcounted by a factor of 3, so there are 16 ways where the middle square is not colored. This leads to a total of 22. |
4277 | 4 | 4 | 8 | Consider an isosceles triangle $T$ with base 10 and height 12. Define a sequence $\omega_{1}, \omega_{2}, \ldots$ of circles such that $\omega_{1}$ is the incircle of $T$ and $\omega_{i+1}$ is tangent to $\omega_{i}$ and both legs of the isosceles triangle for $i>1$. Find the ratio of the radius of $\omega_{i+1}$ to the radius of $\omega_{i}$. | \frac{4}{9} | The ratio of the radius of $\omega_{i+1}$ to the radius of $\omega_{i}$ is $\frac{4}{9}$. |
4294 | 2 | 6 | 8 | Compute the number of nonempty subsets $S \subseteq\{-10,-9,-8, \ldots, 8,9,10\}$ that satisfy $|S|+\min (S)$. $\max (S)=0$. | 335 | Since $\min (S) \cdot \max (S)<0$, we must have $\min (S)=-a$ and $\max (S)=b$ for some positive integers $a$ and $b$. Given $a$ and $b$, there are $|S|-2=a b-2$ elements left to choose, which must come from the set $\{-a+1,-a+2, \ldots, b-2, b-1\}$, which has size $a+b-1$. Therefore the number of possibilities for a given $a, b$ are $\binom{a+b-1}{a b-2}$. In most cases, this binomial coefficient is zero. In particular, we must have $a b-2 \leq a+b-1 \Longleftrightarrow (a-1)(b-1) \leq 2$. This narrows the possibilities for $(a, b)$ to $(1, n)$ and $(n, 1)$ for positive integers $2 \leq n \leq 10$ (the $n=1$ case is impossible), and three extra possibilities: $(2,2),(2,3)$, and $(3,2)$. In the first case, the number of possible sets is $$2\left(\binom{2}{0}+\binom{3}{1}+\cdots+\binom{10}{8}\right)=2\left(\binom{2}{2}+\binom{3}{2}+\cdots+\binom{10}{2}\right)=2\binom{11}{3}=330$$ In the second case the number of possible sets is $$\binom{3}{2}+\binom{4}{4}+\binom{4}{4}=5$$ Thus there are 335 sets in total. |
4295 | 2 | 6 | 8 | $A B C D$ is a regular tetrahedron of volume 1. Maria glues regular tetrahedra $A^{\prime} B C D, A B^{\prime} C D$, $A B C^{\prime} D$, and $A B C D^{\prime}$ to the faces of $A B C D$. What is the volume of the tetrahedron $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$? | \frac{125}{27} | Consider the tetrahedron with vertices at $W=(1,0,0), X=(0,1,0), Y=(0,0,1)$, and $Z=(1,1,1)$. This tetrahedron is similar to $A B C D$. It has center $O=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$. We can construct a tetrahedron $W^{\prime} X^{\prime} Y^{\prime} Z^{\prime}$ in the same way that $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ was constructed by letting $W^{\prime}$ be the reflection of $W$ across $X Y Z$ and so forth. Then we see that $Z^{\prime}=\left(-\frac{1}{3},-\frac{1}{3},-\frac{1}{3}\right)$, so $O Z^{\prime}$ has length $\frac{5}{6} \sqrt{3}$, whereas $O Z$ has length $\frac{1}{2} \sqrt{3}$. We thus see that $W^{\prime} X^{\prime} Y^{\prime} Z^{\prime}$ has a side length $\frac{5}{\frac{6}{2}}=\frac{5}{3}$ that of $W X Y Z$, so by similarity the same is true of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ and $A B C D$. In particular, the volume of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is $\left(\frac{5}{3}\right)^{3}$ that of $A B C D$, so it is $\frac{125}{27}$. |
4297 | 3 | 5 | 8 | Let $A B C$ be a triangle with $A B=A C=5$ and $B C=6$. Denote by $\omega$ the circumcircle of $A B C$. We draw a circle $\Omega$ which is externally tangent to $\omega$ as well as to the lines $A B$ and $A C$ (such a circle is called an $A$-mixtilinear excircle). Find the radius of $\Omega$. | \frac{75}{8} | Let $M$ be the midpoint of $B C$. Let $D$ be the point diametrically opposite $A$ on the circumcircle, and let the $A$-mixtilinear excircle be tangent to lines $A B$ and $A C$ at $X$ and $Y$. Let $O$ be the center of the $A$-mixtilinear excircle. Notice that $\triangle A O X \sim \triangle A B M$. If we let $x$ be the desired radius, we have $$\frac{x+A D}{x}=\frac{5}{3}$$ We can compute $\frac{A D}{5}=\frac{5}{4}$ since $\triangle A D B \sim \triangle A B M$, we derive $A D=\frac{25}{4}$. From here it follows that $x=\frac{75}{8}$. |
4300 | 1 | 7 | 8 | The cells of a $5 \times 5$ grid are each colored red, white, or blue. Sam starts at the bottom-left cell of the grid and walks to the top-right cell by taking steps one cell either up or to the right. Thus, he passes through 9 cells on his path, including the start and end cells. Compute the number of colorings for which Sam is guaranteed to pass through a total of exactly 3 red cells, exactly 3 white cells, and exactly 3 blue cells no matter which route he takes. | 1680 | Let $c_{i, j}$ denote the cell in the $i$-th row from the bottom and the $j$-th column from the left, so Sam starts at $c_{1,1}$ and is traveling to $c_{5,5}$. The key observation (from, say, trying small cases) is that Claim. For $1 \leq i, j<5$, the cells $c_{i+1, j}$ and $c_{i, j+1}$ must be the same color. Proof. Choose a path $P$ from $c_{1,1}$ to $c_{i, j}$, and a path $Q$ from $c_{i+1, j+1}$ to $c_{5,5}$. Then consider the two paths $P \rightarrow c_{i+1, j} \rightarrow Q$ and $P \rightarrow c_{i, j+1} \rightarrow Q$. These both must have 3 cells of each color, but they only differ at cells $c_{i+1, j}$ and $c_{i, j+1}$. So these cells must be the same color. Hence, every diagonal $D_{k}=\left\{c_{a, b}: a+b=k\right\}$ must consist of cells of the same color. Moreover, any path that goes from $c_{1,1}$ to $c_{5,5}$ contains exactly one cell in $D_{k}$ for $k=2,3, \ldots, 10$. So we simply need to color the diagonals $D_{2}, \ldots, D_{10}$ such that there are 3 diagonals of each color. The number of ways to do this is $\binom{9}{3,3,3}=1680$. |
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