The vertices of a regular hexagon are labeled $\cos (\theta), \cos (2 \theta), \ldots, \cos (6 \theta)$. For every pair of vertices, Bob draws a blue line through the vertices if one of these functions can be expressed as a polynomial function of the other (that holds for all real $\theta$ ), and otherwise Roberta draws a red line through the vertices. In the resulting graph, how many triangles whose vertices lie on the hexagon have at least one red and at least one blue edge?

Solution: The existence of the Chebyshev polynomials, which express $\cos (n \theta)$ as a polynomial in $\cos (\theta)$, imply that Bob draws a blue line between $\cos (\theta)$ and each other vertex, and also between $\cos (2 \theta)$ and $\cos (4 \theta)$, between $\cos (2 \theta)$ and $\cos (6 \theta)$, and between $\cos (3 \theta)$ and $\cos (6 \theta)$ (by substituting $\theta^{\prime}=2 \theta$ or $3 \theta$ as necessary). We now show that Roberta draws a red line through each other pair of vertices. Let $m$ and $n$ be positive integers. Notice that $\cos (n \theta)$ is a periodic function with period $\frac{2 \pi}{n}$, and $\cos (m \theta)$ is periodic with period $\frac{2 \pi}{m}$. Thus, any polynomial in $\cos (m \theta)$ is also periodic of period $\frac{2 \pi}{m}$. This may not be the minimum period of the polynomial, however, so the minimum period is $\frac{2 \pi}{m k}$ for some $k$. Therefore, if $\cos (n \theta)$ can be expressed as a polynomial in $\cos (m \theta)$ then $\frac{2 \pi}{n}=\frac{2 \pi}{m k}$ for some $k$, so $m \mid n$. This shows that there is a blue line between two vertices $\cos (a \theta)$ and $\cos (b \theta)$ if and only if one of $a$ or $b$ divides the other. Drawing the graph, one can easily count that there are 3 triangles with all blue edges, 3 triangles with all red edges, and $\binom{6}{3}=20$ triangles total. Thus there are $20-3-3=14$ triangles having at least one red and at least one blue edge.

Answer: 14


Mathematics -> Algebra -> Intermediate Algebra -> Other

Let $a_{1}=1$, and let $a_{n}=\left\lfloor n^{3} / a_{n-1}\right\rfloor$ for $n>1$. Determine the value of $a_{999}$.

Solution: We claim that for any odd $n, a_{n}=n$. The proof is by induction. To get the base cases $n=1$, 3, we compute $a_{1}=1, a_{2}=\left\lfloor 2^{3} / 1\right\rfloor=8, a_{3}=\left\lfloor 3^{3} / 8\right\rfloor=3$. And if the claim holds for odd $n \geq 3$, then $a_{n+1}=\left\lfloor(n+1)^{3} / n\right\rfloor=n^{2}+3 n+3$, so $a_{n+2}=\left\lfloor(n+2)^{3} /\left(n^{2}+3 n+3\right)\right\rfloor=\left\lfloor\left(n^{3}+6 n^{2}+12 n+8\right) /\left(n^{2}+3 n+2\right)\right\rfloor=\left\lfloor n+2+\frac{n^{2}+3 n+2}{n^{2}+3 n+3}\right\rfloor=n+2$. So the claim holds, and in particular, $a_{999}=999$.

Answer: 999


Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations

In the diagram below, how many distinct paths are there from January 1 to December 31, moving from one adjacent dot to the next either to the right, down, or diagonally down to the right?

Solution: For each dot in the diagram, we can count the number of paths from January 1 to it by adding the number of ways to get to the dots to the left of it, above it, and above and to the left of it, starting from the topmost leftmost dot. This yields the following numbers of paths: 372.

Answer: 372


Mathematics -> Algebra -> Algebra -> Polynomial Operations

Mathematics -> Algebra -> Abstract Algebra -> Ring Theory

Let $f$ be a polynomial with integer coefficients such that the greatest common divisor of all its coefficients is 1. For any $n \in \mathbb{N}, f(n)$ is a multiple of 85. Find the smallest possible degree of $f$.

Solution: Notice that, if $p$ is a prime and $g$ is a polynomial with integer coefficients such that $g(n) \equiv 0(\bmod p)$ for some $n$, then $g(n+m p)$ is divisible by $p$ as well for any integer multiple $m p$ of $p$. Therefore, it suffices to find the smallest possible degree of a polynomial $f$ for which $f(0), f(1), f(2), \ldots, f(16)$ are divisible by 17 and by 5. There is a polynomial of degree 17 with integer coefficients having $f(0)=f(1)=\cdots=f(16)=0$, namely $f(x)=(x)(x-1)(x-2) \cdots(x-16)$. Thus the minimal degree is no larger than 17. Now, let $f$ be such a polynomial and consider $f$ modulo 17. The polynomial has 17 roots, so it must be at least degree 17 when taken modulo 17. Thus $f$ has degree at least 17 as well.

Answer: 17


Mathematics -> Discrete Mathematics -> Combinatorics

Mathematics -> Geometry -> Plane Geometry -> Other

Consider a $100 \times 100$ table, and identify the cell in row $a$ and column $b, 1 \leq a, b \leq 100$, with the ordered pair $(a, b)$. Let $k$ be an integer such that $51 \leq k \leq 99$. A $k$-knight is a piece that moves one cell vertically or horizontally and $k$ cells to the other direction; that is, it moves from $(a, b)$ to $(c, d)$ such that $(|a-c|,|b-d|)$ is either $(1, k)$ or $(k, 1)$. The $k$-knight starts at cell $(1,1)$, and performs several moves. A sequence of moves is a sequence of cells $(x_{0}, y_{0})=(1,1)$, $(x_{1}, y_{1}),(x_{2}, y_{2}), \ldots,(x_{n}, y_{n})$ such that, for all $i=1,2, \ldots, n, 1 \leq x_{i}, y_{i} \leq 100$ and the $k$-knight can move from $(x_{i-1}, y_{i-1})$ to $(x_{i}, y_{i})$. In this case, each cell $(x_{i}, y_{i})$ is said to be reachable. For each $k$, find $L(k)$, the number of reachable cells.

Solution: Cell $(x, y)$ is directly reachable from another cell if and only if $x-k \geq 1$ or $x+k \leq 100$ or $y-k \geq 1$ or $y+k \leq 100$, that is, $x \geq k+1$ or $x \leq 100-k$ or $y \geq k+1$ or $y \leq 100-k(*)$. Therefore the cells $(x, y)$ for which $101-k \leq x \leq k$ and $101-k \leq y \leq k$ are unreachable. Let $S$ be this set of unreachable cells in this square, namely the square of cells $(x, y), 101-k \leq x, y \leq k$. If condition $(*)$ is valid for both $(x, y)$ and $(x \pm 2, y \pm 2)$ then one can move from $(x, y)$ to $(x \pm 2, y \pm 2)$, if they are both in the table, with two moves: either $x \leq 50$ or $x \geq 51$; the same is true for $y$. In the first case, move $(x, y) \rightarrow(x+k, y \pm 1) \rightarrow(x, y \pm 2)$ or $(x, y) \rightarrow$ $(x \pm 1, y+k) \rightarrow(x \pm 2, y)$. In the second case, move $(x, y) \rightarrow(x-k, y \pm 1) \rightarrow(x, y \pm 2)$ or $(x, y) \rightarrow(x \pm 1, y-k) \rightarrow(x \pm 2, y)$. Hence if the table is colored in two colors like a chessboard, if $k \leq 50$, cells with the same color as $(1,1)$ are reachable. Moreover, if $k$ is even, every other move changes the color of the occupied cell, and all cells are potentially reachable; otherwise, only cells with the same color as $(1,1)$ can be visited. Therefore, if $k$ is even then the reachable cells consists of all cells except the center square defined by $101-k \leq x \leq k$ and $101-k \leq y \leq k$, that is, $L(k)=100^{2}-(2 k-100)^{2}$; if $k$ is odd, then only half of the cells are reachable: the ones with the same color as $(1,1)$, and $L(k)=\frac{1}{2}(100^{2}-(2 k-100)^{2})$.

Answer: L(k)=\begin{cases}100^{2}-(2 k-100)^{2} & \text{if } k \text{ is even} \\ \frac{100^{2}-(2 k-100)^{2}}{2} & \text{if } k \text{ is odd} \end{cases}


Mathematics -> Algebra -> Prealgebra -> Integers

Find the largest positive integer solution of the equation $\left\lfloor\frac{N}{3}\right\rfloor=\left\lfloor\frac{N}{5}\right\rfloor+\left\lfloor\frac{N}{7}\right\rfloor-\left\lfloor\frac{N}{35}\right\rfloor$.

Solution: 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 $N=65$.

Answer: 65


Mathematics -> Geometry -> Plane Geometry -> Triangulations

Mathematics -> Geometry -> Plane Geometry -> Circles

$A B C D$ 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 $M I / I T$.

Solution: 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 $T$.

Answer: \frac{25}{9}


Kelvin the Frog likes numbers whose digits strictly decrease, but numbers that violate this condition in at most one place are good enough. In other words, if $d_{i}$ denotes the $i$ th digit, then $d_{i} \leq d_{i+1}$ for at most one value of $i$. For example, Kelvin likes the numbers 43210, 132, and 3, but not the numbers 1337 and 123. How many 5-digit numbers does Kelvin like?

Solution: Suppose first that no digit violates the constraint; i.e. the digits are in strictly decreasing order. There are $\binom{10}{5}$ ways to choose the digits of the number, and each set of digits can be arranged in exactly one way, so there are $\binom{10}{5}$ such numbers. We now perform casework on which digit violates the constraint. If it is the final digit, the first four digits must be arranged in decreasing order, which there are $\binom{10}{4}$ ways to do. The final digit can then be any digit, but we have overcounted the ones in which the number is in fully decreasing order (this can happen, for example, if the first 4 digits we chose were $5,4,3$, and 2 - a last digit of 1 was already counted in the first case). Therefore, there are $\binom{10}{4}\binom{10}{1}-252$ new numbers in this case. If the offending digit is second from the right, the first 3 digits must be decreasing, as must the last 2 digits. There are $\binom{10}{3}\binom{10}{2}$ ways to do this. As before, we overcount the case where the second digit from the right is not actually an offender, so we again overcount the case where all 5 digits decrease. Hence there are $\binom{10}{3}\binom{10}{2}-252$ new numbers in this case. The case where the third digit is the offender is identical to the previous case, so there are another $\binom{10}{3}\binom{10}{2}-252$ numbers to account for. The final case is when the second digit is the offending digit, in which case there are $\binom{10}{4}$ ways to choose the final 4 digits, but only 9 ways to choose the opening digit (as 0 cannot be a leading digit). Accounting for the usual overcounting, our final answer is $$252+\left[\binom{10}{4}\binom{10}{1}-252\right]+2\left[\binom{10}{3}\binom{10}{2}-252\right]+\left[\binom{10}{4} \cdot 9-252\right]$$ which is easily calculated as 14034.

Answer: 14034


Stan has a stack of 100 blocks and starts with a score of 0, and plays a game in which he iterates the following two-step procedure: (a) Stan picks a stack of blocks and splits it into 2 smaller stacks each with a positive number of blocks, say $a$ and $b$. (The order in which the new piles are placed does not matter.) (b) Stan adds the product of the two piles' sizes, $a b$, to his score. The game ends when there are only 1-block stacks left. What is the expected value of Stan's score at the end of the game?

Solution: Let $E(n)$ be the expected value of the score for an $n$-block game. It suffices to show that the score is invariant regardless of how the game is played. We proceed by induction. We have $E(1)=0$ and $E(2)=1$. We require that $E(n)=E(n-k)+E(k)+(n-k) k$ for all $k$. Setting $k=1$, we hypothesize that $E(n)=n(n-1) / 2$. This satisfies the recursion and base cases so $E(100)=100 \cdot 99 / 2=4950$.

Answer: 4950


Mathematics -> Algebra -> Algebra -> Sequences and Series

Solution: Let $X$ denote the desired sum. Note that $$\begin{array}{rlr} X & = & \frac{1}{4^{2}}+\frac{1}{4^{3}}+\frac{2}{4^{4}}+\frac{3}{4^{5}}+\frac{5}{4^{6}}+\ldots \\ 4 X & =\quad \frac{1}{4^{1}}+\frac{1}{4^{2}}+\frac{2}{4^{3}}+\frac{3}{4^{4}}+\frac{5}{4^{5}}+\frac{8}{4^{6}}+\ldots \\ 16 X & =\frac{1}{4^{0}}+\frac{1}{4^{1}}+\frac{2}{4^{2}}+\frac{3}{4^{3}}+\frac{5}{4^{4}}+\frac{8}{4^{5}}+\frac{13}{4^{6}}+\ldots \end{array}$$ so that $X+4 X=16 X-1$, and $X=1 / 11$.

Answer: \frac{1}{11}


A candy company makes 5 colors of jellybeans, which come in equal proportions. If I grab a random sample of 5 jellybeans, what is the probability that I get exactly 2 distinct colors?

Solution: There are $\binom{5}{2}=10$ possible pairs of colors. Each pair of colors contributes $2^{5}-2=30$ sequences of beans that use both colors. Thus, the answer is $10 \cdot 30 / 5^{5}=12 / 125$.

Answer: \frac{12}{125}


Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other

Forty two cards are labeled with the natural numbers 1 through 42 and randomly shuffled into a stack. One by one, cards are taken off of the top of the stack until a card labeled with a prime number is removed. How many cards are removed on average?

Solution: Note that there are 13 prime numbers amongst the cards. We may view these as separating the remaining 29 cards into 14 groups of nonprimes - those appearing before the first prime, between the first and second, etc. Each of these groups is equally likely to appear first, so 29/14 nonprimes are removed on average. We are done since exactly one prime is always drawn.

Answer: \frac{43}{14}


Suppose $a, b$, and $c$ are distinct positive integers such that $\sqrt{a \sqrt{b \sqrt{c}}}$ is an integer. Compute the least possible value of $a+b+c$.

Solution: First, check that no permutation of $(1,2,3)$ works, so the sum must be more than 6 . Then since $(a, b, c)=(2,4,1)$ has $\sqrt{2 \sqrt{4 \sqrt{1}}}=2$, the answer must be $2+4+1=7$.

Answer: 7


Mathematics -> Algebra -> Prealgebra -> Simple Equations

Let $A_{12}$ denote the answer to problem 12. There exists a unique triple of digits $(B, C, D)$ such that $10>A_{12}>B>C>D>0$ and $$\overline{A_{12} B C D}-\overline{D C B A_{12}}=\overline{B D A_{12} C}$$ where $\overline{A_{12} B C D}$ denotes the four digit base 10 integer. Compute $B+C+D$.

Solution: Since $D<A_{12}$, when $A$ is subtracted from $D$ we must carry over from $C$. Thus, $D+10-A_{12}=C$. Next, since $C-1<C<B$, we must carry over from the tens digit, so that $(C-1+10)-B=A_{12}$. Now $B>C$ so $B-1 \geq C$, and $(B-1)-C=D$. Similarly, $A_{12}-D=B$. Solving this system of four equations produces $\left(A_{12}, B, C, D\right)=(7,6,4,1)$.

Answer: 11


Mathematics -> Geometry -> Plane Geometry -> Polygons

Compute the circumradius of cyclic hexagon $A B C D E F$, which has side lengths $A B=B C=$ $2, C D=D E=9$, and $E F=F A=12$.

Solution: Construct point $E^{\prime}$ on the circumcircle of $A B C D E F$ such that $D E^{\prime}=E F=12$ and $E^{\prime} F=D E=9$; then $\overline{B E^{\prime}}$ is a diameter. Let $B E^{\prime}=d$. Then $C E^{\prime}=\sqrt{B E^{\prime 2}-B C^{2}}=\sqrt{d^{2}-4}$ and $B D=\sqrt{B E^{\prime 2}-D E^{\prime 2}}=\sqrt{d^{2}-144}$. Applying Ptolemy's theorem to $B C D E^{\prime}$ now yields $$9 \cdot d+2 \cdot 12=\sqrt{\left(d^{2}-4\right)\left(d^{2}-144\right)}$$ Squaring and rearranging, we find $0=d^{4}-229 d^{2}-432 d=d(d-16)\left(d^{2}+16 d+27\right)$. Since $d$ is a positive real number, $d=16$, and the circumradius is 8.

Answer: 8


Mathematics -> Geometry -> Plane Geometry -> Angles

Convex quadrilateral $A B C D$ has right angles $\angle A$ and $\angle C$ and is such that $A B=B C$ and $A D=C D$. The diagonals $A C$ and $B D$ intersect at point $M$. Points $P$ and $Q$ lie on the circumcircle of triangle $A M B$ and segment $C D$, respectively, such that points $P, M$, and $Q$ are collinear. Suppose that $m \angle A B C=160^{\circ}$ and $m \angle Q M C=40^{\circ}$. Find $M P \cdot M Q$, given that $M C=6$.

Solution: Note that $m \angle Q P B=m \angle M P B=m \angle M A B=m \angle C A B=\angle B C A=\angle C D B$. Thus, $M P \cdot M Q=M B \cdot M D$. On the other hand, segment $C M$ is an altitude of right triangle $B C D$, so $M B \cdot M D=M C^{2}=36$.

Answer: 36


Mathematics -> Algebra -> Algebra -> Equations and Inequalities

Suppose $m>n>1$ are positive integers such that there exist $n$ complex numbers $x_{1}, x_{2}, \ldots, x_{n}$ for which - $x_{1}^{k}+x_{2}^{k}+\cdots+x_{n}^{k}=1$ for $k=1,2, \ldots, n-1$ - $x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}=2$; and - $x_{1}^{m}+x_{2}^{m}+\cdots+x_{n}^{m}=4$. Compute the smallest possible value of $m+n$.

Solution: Let $S_{k}=\sum_{j=1}^{n} x_{j}^{k}$, so $S_{1}=S_{2}=\cdots=S_{n-1}=1, S_{n}=2$, and $S_{m}=4$. The first of these conditions gives that $x_{1}, \ldots, x_{n}$ are the roots of $P(x)=x^{n}-x^{n-1}-c$ for some constant $c$. Then $x_{i}^{n}=x_{i}^{n-1}+c$, and thus $$2=S_{n}=S_{n-1}+c n=1+c n$$ so $c=\frac{1}{n}$. Thus, we have the recurrence $S_{k}=S_{k-1}+\frac{S_{k-n}}{n}$. This gives $S_{n+j}=2+\frac{j}{n}$ for $0 \leq j \leq n-1$, and then $S_{2 n}=3+\frac{1}{n}$. Then $S_{2 n+j}=3+\frac{2 j+1}{n}+\frac{j^{2}+j}{2 n^{2}}$ for $0 \leq j \leq n-1$. In particular, $S_{3 n-1}>4$, so we have $m \in[2 n, 3 n-1]$. Let $m=2 n+j$. Then $$3+\frac{2 j+1}{n}+\frac{j^{2}+j}{2 n^{2}}=4 \Longrightarrow 2 n^{2}-2 n(2 j+1)-\left(j^{2}+j\right)=0$$ Viewing this as a quadratic in $n$, the discriminant $4(2 j+1)^{2}+8\left(j^{2}+j\right)=24 j^{2}+24 j+4=4\left(6 j^{2}+6 j+1\right)$ must be a perfect square, so $6 j^{2}+6 j+1$ is a square. Then $$6 j^{2}+6 j+1=y^{2} \Longrightarrow 12 j^{2}+12 j+2=2 y^{2} \Longrightarrow 3(2 j+1)^{2}-2 y^{2}=1$$ The case $j=0$ gives $n=1$, a contradiction. After this, the smallest $j$ that works is $j=4$ (and $y=11$ ). Plugging this back into our quadratic, $$2 n^{2}-18 n-20=0 \Longrightarrow n^{2}-9 n-10=0$$ so $n=10$. Then $m=2 n+j=24$, so $m+n=34$.

Answer: 34


Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Other

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?

Solution: 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 $$\frac{1+\left(1-\frac{1}{1515}\right)\left(1-\frac{1}{1414}\right) \cdots\left(1-\frac{1}{22}\right)}{2}=\frac{1+\frac{1416}{15151315} \cdots \frac{13}{22}}{2}=\frac{1+\frac{8}{15}}{2}=\frac{23}{30}$$

Answer: \frac{23}{30}


Elisenda has a piece of paper in the shape of a triangle with vertices $A, B$, and $C$ such that $A B=42$. She chooses a point $D$ on segment $A C$, and she folds the paper along line $B D$ so that $A$ lands at a point $E$ on segment $B C$. Then, she folds the paper along line $D E$. When she does this, $B$ lands at the midpoint of segment $D C$. Compute the perimeter of the original unfolded triangle.

Solution: Let $F$ be the midpoint of segment $D C$. Evidently $\angle A D B=60^{\circ}=\angle B D E=\angle E D C$. Moreover, we have $B D=D F=F C, A D=D E$, and $A B=B E$. Hence angle bisector on $B D C$ gives us that $B E=42, E C=84$, and hence angle bisector on $A B C$ gives us that if $A D=x$ then $C D=3 x$. Now this gives $B D=3 x / 2$, so thus the Law of Cosines on $A D B$ gives $x=12 \sqrt{7}$. Hence, $B C=42+84=126$ and $A C=4 x=48 \sqrt{7}$. The answer is $42+126+48 \sqrt{7}=168+48 \sqrt{7}$.

Answer: 168+48 \sqrt{7}


Given a positive integer $k$, let \|k\| denote the absolute difference between $k$ and the nearest perfect square. For example, \|13\|=3 since the nearest perfect square to 13 is 16. Compute the smallest positive integer $n$ such that $\frac{\|1\|+\|2\|+\cdots+\|n\|}{n}=100$.

Solution: Note that from $n=m^{2}$ to $n=(m+1)^{2},\|n\|$ increases from 0 to a peak of $m$ (which is repeated twice), and then goes back down to 0. Therefore $\sum_{n=1}^{m^{2}}\|n\|=\sum_{k=1}^{m-1} 2(1+2+\cdots+k)=\sum_{k=1}^{m-1} 2\binom{k+1}{2}=2\binom{m+1}{3}=\frac{m}{3}\left(m^{2}-1\right)$. In particular, if $n=m^{2}-1$, $\frac{\|1\|+\|2\|+\cdots+\|n\|}{n}=\frac{m}{3}$ so $n=300^{2}-1$ satisfies the condition. However, this does not prove that there are not smaller solutions for $n$. Let $N=300^{2}-1$ and suppose that $N-k$ satisfies the condition. Then, we know that $\frac{\|N\|+\|N-1\|+\cdots\|N-(k-1)\|}{k}=100$. Since \|N-k\|=k+1 for $k \leq 298$, one can show that $k=199$ works. By looking at further terms, one can convince oneself that no larger value of $k$ works. Thus, the answer is $300^{2}-1-199=90000-200=$ 89800.

Answer: 89800


The graph of the equation $x+y=\left\lfloor x^{2}+y^{2}\right\rfloor$ consists of several line segments. Compute the sum of their lengths.

Solution: We split into cases on the integer $k=\left\lfloor x^{2}+y^{2}\right\rfloor$. Note that $x+y=k$ but $x^{2}+y^{2} \geq$ $\frac{1}{2}(x+y)^{2}=\frac{1}{2} k^{2}$ and $x^{2}+y^{2}<k+1$, which forces $k \leq 2$. If $k=0$, the region defined by $0 \leq x^{2}+y^{2}<1$ and $x+y=0$ is the diameter from $\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$ to $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, which has length 2 . If $k=1$, the region $1 \leq x^{2}+y^{2}<2$ and $x+y=1$ consists of two segments, which is the chord on $x^{2}+y^{2}=2$ minus the chord on $x^{2}+y^{2}=1$. The former has length $2 \sqrt{(\sqrt{2})^{2}-\left(\frac{\sqrt{2}}{2}\right)^{2}}=\sqrt{6}$, and the latter has length $2 \sqrt{1^{2}-\left(\frac{\sqrt{2}}{2}\right)^{2}}=\sqrt{2}$. So the total length here is $\sqrt{6}-\sqrt{2}$. If $k=2$, the region $2 \leq x^{2}+y^{2}<3$ and $x+y=1$ is the chord on $x^{2}+y^{2}=3$, which has length $2 \sqrt{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=2$. Our final answer is $2+(\sqrt{6}-\sqrt{2})+2=4+\sqrt{6}-\sqrt{2}$.

Answer: 4+\sqrt{6}-\sqrt{2}


Find the number of 7 -tuples $\left(n_{1}, \ldots, n_{7}\right)$ of integers such that $$\sum_{i=1}^{7} n_{i}^{6}=96957$$

Solution: Consider the equation in modulo 9. All perfect 6 th powers are either 0 or 1. Since 9 divides 96957, it must be that each $n_{i}$ is a multiple of 3. Writing $3 a_{i}=n_{i}$ and dividing both sides by $3^{6}$, we have $a_{1}^{6}+\cdots+a_{7}^{6}=133$. Since sixth powers are nonnegative, $\left|a_{i}\right| \leq 2$. Again considering modulo 9, we see that $a_{i} \neq 0$. Thus, $a_{i}^{6} \in\{1,64\}$. The only possibility is $133=64+64+1+1+1+1+1$, so $\left|a_{1}\right|, \ldots,\left|a_{7}\right|$ consists of 22 's and 51 's. It follows that the answer is $\binom{7}{2} \cdot 2^{7}=2688$.

Answer: 2688


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 $A T \cdot B T$.

Solution: 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 $A T \cdot B T=T P^{2}-A B \cdot T P \cdot M O / A M=14^{2}-24 \cdot 14 \cdot 5 / 12=56$.

Answer: 56


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.

Solution: 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.

Answer: 32


Points $A, B$, and $C$ lie in that order on line $\ell$, such that $A B=3$ and $B C=2$. Point $H$ is such that $C H$ is perpendicular to $\ell$. Determine the length $C H$ such that $\angle A H B$ is as large as possible.

Solution: Let $\omega$ denote the circumcircle of triangle $A B H$. Since $A B$ is fixed, the smaller the radius of $\omega$, the bigger the angle $A H B$. If $\omega$ crosses the line $C H$ in more than one point, then there exists a smaller circle that goes through $A$ and $B$ that crosses $C H$ at a point $H^{\prime}$. But angle $A H^{\prime} B$ is greater than $A H B$, contradicting our assumption that $H$ is the optimal spot. Thus the circle $\omega$ crosses the line $C H$ at exactly one spot: ie, $\omega$ is tangent to $C H$ at $H$. By Power of a Point, $C H^{2}=C A C B=5 \cdot 2=10$, so $C H=\sqrt{10}$.

Answer: \sqrt{10}


Mathematics -> Number Theory -> Prime Numbers

How many positive integers less than 1998 are relatively prime to 1547 ? (Two integers are relatively prime if they have no common factors besides 1.)

Solution: The factorization of 1547 is \(7 \cdot 13 \cdot 17\), so we wish to find the number of positive integers less than 1998 that are not divisible by 7, 13, or 17. By the Principle of Inclusion-Exclusion, we first subtract the numbers that are divisible by one of 7, 13, and 17, add back those that are divisible by two of 7, 13, and 17, then subtract those divisible by three of them. That is, \(1997-\left\lfloor\frac{1997}{7}\right\rfloor-\left\lfloor\frac{1997}{13}\right\rfloor-\left\lfloor\frac{1997}{17}\right\rfloor+\left\lfloor\frac{1997}{7 \cdot 13}\right\rfloor+\left\lfloor\frac{1997}{7 \cdot 17}\right\rfloor+\left\lfloor\frac{1997}{13 \cdot 17}\right\rfloor-\left\lfloor\frac{1997}{7 \cdot 13 \cdot 17}\right\rfloor\) or 1487.

Answer: 1487


Compute, in terms of $n$, $\sum_{k=0}^{n}\binom{n-k}{k} 2^{k}$.

Solution: Let $T_{n}=\sum_{k=0}^{n}\binom{n-k}{k} 2^{k}$. From Pascal's recursion for binomial coefficients, we can find $T_{n}=2 T_{n-2}+T_{n-1}$, with $T_{0}=1$ and $T_{1}=1$. The characteristic polynomial of this recursion is $x^{2}-x-2=0$, which has roots 2 and -1. Thus $T_{n}=a \cdot 2^{n}+b \cdot(-1)^{n}$ for some $a$ and $b$. From the initial conditions we have $a+b=1$ and $2 a-b=1$. It follows that $a=2 / 3$ and $b=1 / 3$, from which the conclusion follows.

Answer: \frac{2 \cdot 2^{n}+(-1)^{n}}{3}


Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers

Kelvin and 15 other frogs are in a meeting, for a total of 16 frogs. During the meeting, each pair of distinct frogs becomes friends with probability $\frac{1}{2}$. Kelvin thinks the situation after the meeting is cool if for each of the 16 frogs, the number of friends they made during the meeting is a multiple of 4. Say that the probability of the situation being cool can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are relatively prime. Find $a$.

Solution: Consider the multivariate polynomial $$\prod_{1 \leq i<j \leq 16}\left(1+x_{i} x_{j}\right)$$ We're going to filter this by summing over all $4^{16} 16$-tuples $\left(x_{1}, x_{2}, \ldots, x_{16}\right)$ such that $x_{j}= \pm 1, \pm i$. Most of these evaluate to 0 because $i^{2}=(-i)^{2}=-1$, and $1 \cdot-1=-1$. If you do this filtering, you get the following 4 cases: Case 1: Neither of $i$ or $-i$ appears. Then the only cases we get are when all the $x_{j}$ are 1, or they're all -1. Total is $2^{121}$. Case 2: $i$ appears, but $-i$ does not. Then all the remaining $x_{j}$ must be all 1 or all -1. This contributes a sum of $(1+i)^{15} \cdot 2^{105}+(1-i)^{15} \cdot 2^{105}=2^{113}$. $i$ can be at any position, so we get $16 \cdot 2^{113}$. Case 3: $-i$ appears, but $i$ does not. Same contribution as above. $16 \cdot 2^{113}$. Case 4: Both $i$ and $-i$ appear. Then all the rest of the $x_{j}$ must be all 1 or all -1. This contributes a sum of $2 \cdot(1+i(-i)) \cdot(1+i)^{14} \cdot(1-i)^{14} \cdot 2^{91}=2^{107}$. $i$ and $-i$ can appear in $16 \cdot 15$ places, so we get $240 \cdot 2^{107}$. So the final answer is this divided a factor for our filter. $\left(4^{16}=2^{32}\right.$.) So our final answer is $\frac{2^{89}+16 \cdot 2^{82}+240 \cdot 2^{75}}{2^{120}}=\frac{1167}{2^{41}}$. Therefore, the answer is 1167.

Answer: 1167


Mathematics -> Number Theory -> Congruences

Mathematics -> Algebra -> Algebra -> Polynomial Operations

Find all positive integers $n$ such that for any integer $k$ there exists an integer $a$ for which $a^{3}+a-k$ is divisible by $n$.

Solution: Answer: All integers $n=3^{b}$, where $b$ is a nonnegative integer. We are looking for integers $n$ such that the set $A=\left\{a^{3}+a \mid a \in \mathbf{Z}\right\}$ is a complete residue system by modulo $n$. Let us call this property by $\left(^{*}\right)$. It is not hard to see that $n=1$ satisfies $\left({ }^{*}\right)$ and $n=2$ does not. If $a \equiv b(\bmod n)$, then $a^{3}+a \equiv b^{3}+b(\bmod n)$. So $n$ satisfies $\left(^{*}\right)$ iff there are no $a, b \in\{0, \ldots, n-1\}$ with $a \neq b$ and $a^{3}+a \equiv b^{3}+b(\bmod n)$. First, let us prove that $3^{j}$ satisfies $\left(^{*}\right)$ for all $j \geq 1$. Suppose that $a^{3}+a \equiv b^{3}+b\left(\bmod 3^{j}\right)$ for $a \neq b$. Then $(a-b)\left(a^{2}+a b+b^{2}+1\right) \equiv 0\left(\bmod 3^{j}\right)$. We can easily check $\bmod 3$ that $a^{2}+a b+b^{2}+1$ is not divisible by 3 . Next note that if $A$ is not a complete residue system modulo integer $r$, then it is also not a complete residue system modulo any multiple of $r$. Hence it remains to prove that any prime $p>3$ does not satisfy $\left({ }^{*}\right)$. If $p \equiv 1(\bmod 4)$, there exists $b$ such that $b^{2} \equiv-1(\bmod p)$. We then take $a=0$ to obtain the congruence $a^{3}+a \equiv b^{3}+b(\bmod p)$. Suppose now that $p \equiv 3(\bmod 4)$. We will prove that there are integers $a, b \not \equiv 0(\bmod p)$ such that $a^{2}+a b+b^{2} \equiv-1(\bmod p)$. Note that we may suppose that $a \not \equiv b(\bmod p)$, since otherwise if $a \equiv b(\bmod p)$ satisfies $a^{2}+a b+b^{2}+1 \equiv 0(\bmod p)$, then $(2 a)^{2}+(2 a)(-a)+$ $a^{2}+1 \equiv 0(\bmod p)$ and $2 a \not \equiv-a(\bmod p)$. Letting $c$ be the inverse of $b$ modulo $p$ (i.e. $b c \equiv 1(\bmod p))$, the relation is equivalent to $(a c)^{2}+a c+1 \equiv-c^{2}(\bmod p)$. Note that $-c^{2}$ can take on the values of all non-quadratic residues modulo $p$. If we can find an integer $x$ such that $x^{2}+x+1$ is a non-quadratic residue modulo $p$, the values of $a$ and $c$ will follow immediately. Hence we focus on this latter task. Note that if $x, y \in\{0, \ldots, p-1\}=B$, then $x^{2}+x+1 \equiv y^{2}+y+1(\bmod p)$ iff $p$ divides $x+y+1$. We can deduce that $x^{2}+x+1$ takes on $(p+1) / 2$ values as $x$ varies in $B$. Since there are $(p-1) / 2$ non-quadratic residues modulo $p$, the $(p+1) / 2$ values that $x^{2}+x+1$ take on must be 0 and all the quadratic residues. Let $C$ be the set of quadratic residues modulo $p$ and 0 , and let $y \in C$. Suppose that $y \equiv z^{2}(\bmod p)$ and let $z \equiv 2 w+1(\bmod p)$ (we can always choose such $\left.w\right)$). Then $y+3 \equiv$ $4\left(w^{2}+w+1\right)(\bmod p)$. From the previous paragraph, we know that $4\left(w^{2}+w+1\right) \in C$. This means that $y \in C \Longrightarrow y+3 \in C$. Unless $p=3$, the relation implies that all elements of $B$ are in $C$, a contradiction. This concludes the proof.

Answer: All integers $n=3^{b}$, where $b$ is a nonnegative integer.


$A B C D$ is a cyclic quadrilateral in which $A B=4, B C=3, C D=2$, and $A D=5$. Diagonals $A C$ and $B D$ intersect at $X$. A circle $\omega$ passes through $A$ and is tangent to $B D$ at $X . \omega$ intersects $A B$ and $A D$ at $Y$ and $Z$ respectively. Compute $Y Z / B D$.

Solution: Denote the lengths $A B, B C, C D$, and $D A$ by $a, b, c$, and $d$ respectively. Because $A B C D$ is cyclic, $\triangle A B X \sim \triangle D C X$ and $\triangle A D X \sim \triangle B C X$. It follows that $\frac{A X}{D X}=\frac{B X}{C X}=\frac{a}{c}$ and $\frac{A X}{B X}=\frac{D X}{C X}=\frac{d}{b}$. Therefore we may write $A X=a d k, B X=a b k, C X=b c k$, and $D X=c d k$ for some $k$. Now, $\angle X D C=\angle B A X=\angle Y X B$ and $\angle D C X=\angle X B Y$, so $\triangle B X Y \sim \triangle C D X$. Thus, $X Y=$ $D X \cdot \frac{B X}{C D}=a b d k^{2}$. Analogously, $X Y=a c d k^{2}$. Note that $X Y / X Z=C B / C D$. Since $\angle Y X Z=\pi-\angle Z A \stackrel{c}{Y}=\angle B C D$, we have that $\triangle X Y Z \sim \triangle C B D$. Thus, $Y Z / B D=X Y / C B=a d k^{2}$. Finally, Ptolemy's theorem applied to $A B C D$ gives $$(a d+b c) k \cdot(a b+c d) k=a c+b d$$ It follows that the answer is $$\frac{a d(a c+b d)}{(a b+c d)(a d+b c)}=\frac{20 \cdot 23}{22 \cdot 26}=\frac{115}{143}$$

Answer: \frac{115}{143}


Mathematics -> Applied Mathematics -> Probability -> Other

The spikiness of a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of at least two real numbers is the sum $\sum_{i=1}^{n-1}\left|a_{i+1}-a_{i}\right|$. Suppose $x_{1}, x_{2}, \ldots, x_{9}$ are chosen uniformly and randomly from the interval $[0,1]$. Let $M$ be the largest possible value of the spikiness of a permutation of $x_{1}, x_{2}, \ldots, x_{9}$. Compute the expected value of $M$.

Solution: Our job is to arrange the nine numbers in a way that maximizes the spikiness. Let an element be a peak if it is higher than its neighbor(s) and a valley if it is lower than its neighbor(s). It is not hard to show that an optimal arrangement has every element either a peak or a valley (if you have some number that is neither, just move it to the end to increase spikiness). Since 9 is odd, there are two possibilities: the end points are either both peaks or both valleys. Sort the numbers from least to greatest: $x_{1}, \ldots, x_{9}$. If we arrange them in such a way that it starts and ends with peaks, the factor of $x_{i}$ added to the final result will be $[-2,-2,-2,-2,1,1,2,2,2]$, respectively. If we choose the other way (starting and ending with valleys), we get $[-2,-2,-2,-1,-1,2,2,2,2]$. Notice that both cases have a base value of $[-2,-2,-2,-1,0,1,2,2,2]$, but then we add on $\max \left(x_{6}-\right.$ $\left.x_{5}, x_{5}-x_{4}\right)$. Since the expected value of $x_{i}$ is $\frac{i}{10}$, our answer is $-\frac{2}{10}(1+2+3)-\frac{4}{10}+\frac{6}{10}+\frac{2}{10}(7+8+$ $9)+\mathbb{E}\left(\max \left(x_{6}-x_{5}, x_{5}-x_{4}\right)\right)$. This last term actually has value $\frac{3}{4} \mathbb{E}\left(x_{6}-x_{4}\right)=\frac{3}{4} \cdot \frac{2}{10}$. This is because if we fix all values except $x_{5}$, then $x_{5}$ is uniformly distributed in $\left[x_{4}, x_{6}\right]$. Geometric probability tells us that the distance from $x_{5}$ to its farthest neighbor is $\frac{3}{4}$ to total distance betwen its two neighbors $\left(x_{6}-x_{4}\right)$. We add this all up to get $\frac{79}{20}$.

Answer: \frac{79}{20}


Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations

Von Neumann's Poker: The first step in Von Neumann's game is selecting a random number on $[0,1]$. To generate this number, Chebby uses the factorial base: the number $0 . A_{1} A_{2} A_{3} A_{4} \ldots$ stands for $\sum_{n=0}^{\infty} \frac{A_{n}}{(n+1)!}$, where each $A_{n}$ is an integer between 0 and $n$, inclusive. Chebby has an infinite number of cards labeled $0, 1, 2, \ldots$. He begins by putting cards $0$ and $1$ into a hat and drawing randomly to determine $A_{1}$. The card assigned $A_{1}$ does not get reused. Chebby then adds in card 2 and draws for $A_{2}$, and continues in this manner to determine the random number. At each step, he only draws one card from two in the hat. Unfortunately, this method does not result in a uniform distribution. What is the expected value of Chebby's final number?

Solution: The expected value of a number generated in the factorial base system is given by the sum of the expected values of each digit divided by its factorial weight. For this specific setup, the expected value is approximately 0.57196.

Answer: 0.57196


For $a$ a positive real number, let $x_{1}, x_{2}, x_{3}$ be the roots of the equation $x^{3}-a x^{2}+a x-a=0$. Determine the smallest possible value of $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-3 x_{1} x_{2} x_{3}$.

Solution: Note that $x_{1}+x_{2}+x_{3}=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a$. Then $$\begin{aligned} & x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-3 x_{1} x_{2} x_{3}=\left(x_{1}+x_{2}+x_{3}\right)\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)\right) \\ & \quad=\left(x_{1}+x_{2}+x_{3}\right)\left(\left(x_{1}+x_{2}+x_{3}\right)^{2}-3\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)\right)=a \cdot\left(a^{2}-3 a\right)=a^{3}-3 a^{2} \end{aligned}$$ The expression is negative only where $0<a<3$, so we need only consider these values of $a$. Finally, AM-GM gives $\sqrt[3]{(6-2 a)(a)(a)} \leq \frac{(6-2 a)+a+a}{3}=2$, with equality where $a=2$, and this rewrites as $(a-3) a^{2} \geq-4$

Answer: -4


There are three video game systems: the Paystation, the WHAT, and the ZBoz2 \pi, and none of these systems will play games for the other systems. Uncle Riemann has three nephews: Bernoulli, Galois, and Dirac. Bernoulli owns a Paystation and a WHAT, Galois owns a WHAT and a ZBoz2 \pi, and Dirac owns a ZBoz2 \pi and a Paystation. A store sells 4 different games for the Paystation, 6 different games for the WHAT, and 10 different games for the ZBoz2 \pi. Uncle Riemann does not understand the difference between the systems, so he walks into the store and buys 3 random games (not necessarily distinct) and randomly hands them to his nephews. What is the probability that each nephew receives a game he can play?

Solution: Since the games are not necessarily distinct, probabilities are independent. Multiplying the odds that each nephew receives a game he can play, we get $10 / 20 \cdot 14 / 20 \cdot 16 / 20=7 / 25$.

Answer: \frac{7}{25}


Mathematics -> Number Theory -> Perfect Squares -> Other

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.

Solution: 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.

Answer: (-4,-4),(-5,-6),(-6,-5),(0, k^{2}),(k^{2}, 0),(k, 1-k)


Mathematics -> Algebra -> Intermediate Algebra -> Logarithmic Functions

Suppose a real number \(x>1\) satisfies \(\log _{2}\left(\log _{4} x\right)+\log _{4}\left(\log _{16} x\right)+\log _{16}\left(\log _{2} x\right)=0\). Compute \(\log _{2}\left(\log _{16} x\right)+\log _{16}\left(\log _{4} x\right)+\log _{4}\left(\log _{2} x\right)\).

Solution: Let \(A\) and \(B\) be these sums, respectively. Then \(B-A =\log _{2}\left(\frac{\log _{16} x}{\log _{4} x}\right)+\log _{4}\left(\frac{\log _{2} x}{\log _{16} x}\right)+\log _{16}\left(\frac{\log _{4} x}{\log _{2} x}\right) =\log _{2}\left(\log _{16} 4\right)+\log _{4}\left(\log _{2} 16\right)+\log _{16}\left(\log _{4} 2\right) =\log _{2}\left(\frac{1}{2}\right)+\log _{4} 4+\log _{16}\left(\frac{1}{2}\right) =(-1)+1+\left(-\frac{1}{4}\right) =-\frac{1}{4}\). Since \(A=0\), we have the answer \(B=-\frac{1}{4}\).

Answer: -\frac{1}{4}


Mathematics -> Algebra -> Intermediate Algebra -> Other

For positive integers $n$ and $k$, let $\mho(n, k)$ be the number of distinct prime divisors of $n$ that are at least $k$. Find the closest integer to $$\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{\mho(n, k)}{3^{n+k-7}}$$

Solution: A prime $p$ is counted in $\mho(n, k)$ if $p \mid n$ and $k \leq p$. Thus, for a given prime $p$, the total contribution from $p$ in the sum is $$3^{7} \sum_{m=1}^{\infty} \sum_{k=1}^{p} \frac{1}{3^{p m+k}}=3^{7} \sum_{i \geq p+1} \frac{1}{3^{i}}=\frac{3^{7-p}}{2}$$ Therefore, if we consider $p \in\{2,3,5,7, \ldots\}$ we get $$\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{\mho(n, k)}{3^{n+k-7}}=\frac{3^{5}}{2}+\frac{3^{4}}{2}+\frac{3^{2}}{2}+\frac{3^{0}}{2}+\varepsilon=167+\varepsilon$$ where $\varepsilon<\sum_{i=11}^{\infty} \frac{3^{7-i}}{2}=\frac{1}{108} \ll \frac{1}{2}$. The closest integer to the sum is 167.

Answer: 167


Mathematics -> Applied Mathematics -> Statistics -> Mathematical Statistics

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.

Solution: 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 \(N \geq 34\).

Answer: 34


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?

Solution: 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.

Answer: 12679605


Mathematics -> Geometry -> Plane Geometry -> Area

A square can be divided into four congruent figures as shown: If each of the congruent figures has area 1, what is the area of the square?

Solution: There are four congruent figures with area 1, so the area of the square is 4.

Answer: 4


Mathematics -> Algebra -> Abstract Algebra -> Other

Let $n$ be a positive integer, and let $s$ be the sum of the digits of the base-four representation of $2^{n}-1$. If $s=2023$ (in base ten), compute $n$ (in base ten).

Solution: Every power of 2 is either represented in base 4 as $100 \ldots 00_{4}$ or $200 . .00_{4}$ with some number of zeros. That means every positive integer in the form $2^{n}-1$ is either represented in base 4 as $333 \ldots 33_{4}$ or $133 \ldots 33$ for some number threes. Note that $2023=2022+1=674 \cdot 3+1$, meaning $2^{n}-1$ must be $133 \ldots 333_{4}$ with 674 threes. Converting this to base 2 results in $$133 \ldots 33_{4}=200 \ldots 00_{4}-1=2 \cdot 4^{674}-1=2^{1349}-1$$ for an answer of 1349.

Answer: 1349


Mathematics -> Calculus -> Integral Calculus -> Techniques of Integration -> Single-variable

Mathematics -> Algebra -> Algebraic Expressions -> Other

Solution: Note that $$n^{4}+n^{2}+1=\left(n^{4}+2 n^{2}+1\right)-n^{2}=\left(n^{2}+1\right)^{2}-n^{2}=\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)$$ Decomposing into partial fractions, we find that $$\frac{n}{n^{4}+n^{2}+1}=\frac{1}{2}\left(\frac{1}{n^{2}-n+1}-\frac{1}{n^{2}+n+1}\right)$$ Now, note that if $f(n)=\frac{1}{n^{2}-n+1}$, then $f(n+1)=\frac{1}{(n+1)^{2}-(n+1)+1}=\frac{1}{n^{2}+n+1}$. It follows that $$\sum_{n=0}^{\infty} \frac{n}{n^{4}+n^{2}+1}=\frac{1}{2}((f(0)-f(1))+(f(1)-f(2))+(f(2)-f(3))+\cdots)$$ Since $f(n)$ tends towards 0 as $n$ gets large, this sum telescopes to $f(0) / 2=1 / 2$.

Answer: 1/2


Given that \(x\) is a positive real, find the maximum possible value of \(\sin \left(\tan ^{-1}\left(\frac{x}{9}\right)-\tan ^{-1}\left(\frac{x}{16}\right)\right)\).

Solution: Consider a right triangle \(A O C\) with right angle at \(O, A O=16\) and \(C O=x\). Moreover, let \(B\) be on \(A O\) such that \(B O=9\). Then \(\tan ^{-1} \frac{x}{9}=\angle C B O\) and \(\tan ^{-1} \frac{x}{16}=\angle C A O\), so their difference is equal to \(\angle A C B\). Note that the locus of all possible points \(C\) given the value of \(\angle A C B\) is part of a circle that passes through \(A\) and \(B\), and if we want to maximize this angle then we need to make this circle as small as possible. This happens when \(O C\) is tangent to the circumcircle of \(A B C\), so \(O C^{2}=O A \cdot O B=144=12^{2}\), thus \(x=12\), and it suffices to compute \(\sin (\alpha-\beta)\) where \(\sin \alpha=\cos \beta=\frac{4}{5}\) and \(\cos \alpha=\sin \beta=\frac{3}{5}\). By angle subtraction formula we get \(\sin (\alpha-\beta)=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}=\frac{7}{25}\).

Answer: \frac{7}{25}


Mathematics -> Geometry -> Plane Geometry -> Polygons

Compute the number of integers \(n \in\{1,2, \ldots, 300\}\) such that \(n\) is the product of two distinct primes, and is also the length of the longest leg of some nondegenerate right triangle with integer side lengths.

Solution: Let \(n=p \cdot q\) for primes \(p<q\). If \(n\) is the second largest side of a right triangle there exist integers \(c, a\) such that \(a<p q\) and \((p q)^{2}=c^{2}-a^{2}=(c-a)(c+a)\). Since \(c-a<c+a\) there are three cases for the values of \(c-a, c+a\), and in each case we determine when \(a<p q\). (a) \(c-a=1\) and \(c+a=p^{2} q^{2}\) : Then \(a=\frac{p^{2} q^{2}-1}{2}>p q\), so there are no solutions. (b) \(c-a=p\) and \(c+a=p q^{2}\) : Then \(a=\frac{p q^{2}-p}{2}>p q\). (c) \(c-a=p^{2}\) and \(c+a=q^{2}\). Then \(a=\frac{q^{2}-p^{2}}{2}\) which we require to be less than \(p q\). This is equivalent to \(\frac{q^{2}-p^{2}}{2} <p q q^{2} <2 p q+p^{2} 2 q^{2} <(q+p)^{2} \sqrt{2} q <q+p (\sqrt{2}-1) q <p<q\). So the problem is equivalent to finding the number of distinct prime pairs \((p, q)\) such that \(p q<300\) and \((\sqrt{2}-1) q<p<q\). There are 13 such pairs: \(\{(3,5),(3,7),(5,7),(5,11),(7,11),(7,13),(11,13),(11,17),(11,19),(11,23),(13,17),(13,19),(13,23)\}\) and \(13 \cdot 23=299\) which is the biggest such pair. The most interesting borderline case are \((3,7): \frac{3}{7} \approx .42>\sqrt{2}-1\), which leads to the \((20,21,29)\) triangle, \((5,13): \frac{5}{13} \approx .385<\sqrt{2}-1\), which leads to the \((65,72,97)\) triangle, and \((7,17): \frac{7}{17} \approx .411<\sqrt{2}-1\) which leads to the \((119,120,169)\) right triangle.

Answer: 13


Mathematics -> Geometry -> Plane Geometry -> Triangulations

Find the maximum number of points $X_{i}$ such that for each $i$, $\triangle A B X_{i} \cong \triangle C D X_{i}$.

Solution: One of the sides $A X_{i}$ or $B X_{i}$ is equal to $C D$, thus $X_{i}$ is on one of the circles of radius $C D$ and center $A$ or $B$. In the same way $X_{i}$ is on one of circles of radius $A B$ with center $C$ or $D$. The intersection of these four circles has no more than 8 points so that $n \leq 8$. Suppose that circle $S_{B}$ with center $B$ and radius $C D$ intersects circle $S_{C}$ with center $C$ and radius $A B$ in two points $X_{1}$ and $X_{2}$ which satisfy the conditions of the problem. Then in triangles $A B X_{1}$ and $C D X_{1}$ we have $B X_{1}=C D$ and $C X_{1}=A B$. Since these triangles are congruent then $A X_{1}=D X_{1}$, therefore $X_{1}$ and $X_{2}$ are on the perpendicular bisector of $A D$. On the other hand $X_{1} X_{2}$ is perpendicular to segment $B C$. Then $B C \| A D$ and $A B$ and $C D$ are the diagonals or nonparallel sides of a trapezoid. Suppose that $A B<C D$. Then $B X_{1}=C D>A B=C X_{1}$. It follows that the distance from $A$ to the perpendicular bisector of $B C$ must be less than the distance from $D$ to this line otherwise we obtain a contradiction to the condition $A B<C D$. Then for any point $X$ in the perpendicular bisector of $B C$ we have $A X<D X$ and it is not possible to have $A X=C D, D X=A B$. Thus if the circle with center $A$ and radius $C D$ intersects the circle with center $D$ and radius $A B$, then the points of intersection do not satisfy the condition of congruence. Therefore if the points of intersection of $S_{B}$ with $S_{C}$ satisfy the condition of congruence, then the points of intersection of $S_{A}$ with $S_{D}$ do not. Thus no more than half of the 8 points of intersection of these circles can satisfy the condition of congruence, i.e. $n \leq 4$. If $n=4$ we have the following example of a regular hexagon.

Answer: 4


Mathematics -> Algebra -> Algebra -> Algebraic Expressions

Two real numbers $x$ and $y$ are such that $8 y^{4}+4 x^{2} y^{2}+4 x y^{2}+2 x^{3}+2 y^{2}+2 x=x^{2}+1$. Find all possible values of $x+2 y^{2}$.

Solution: Writing $a=x+2 y^{2}$, the given quickly becomes $4 y^{2} a+2 x^{2} a+a+x=x^{2}+1$. We can rewrite $4 y^{2} a$ for further reduction to $a(2 a-2 x)+2 x^{2} a+a+x=x^{2}+1$, or $$\begin{equation*} 2 a^{2}+\left(2 x^{2}-2 x+1\right) a+\left(-x^{2}+x-1\right)=0 \tag{*} \end{equation*}$$ The quadratic formula produces the discriminant $$\left(2 x^{2}-2 x+1\right)^{2}+8\left(x^{2}-x+1\right)=\left(2 x^{2}-2 x+3\right)^{2}$$ an identity that can be treated with the difference of squares, so that $a=\frac{-2 x^{2}+2 x-1 \pm\left(2 x^{2}-2 x+3\right)}{4}=$ $\frac{1}{2},-x^{2}+x-1$. Now $a$ was constructed from $x$ and $y$, so is not free. Indeed, the second expression flies in the face of the trivial inequality: $a=-x^{2}+x-1<-x^{2}+x \leq x+2 y^{2}=a$. On the other hand, $a=1 / 2$ is a bona fide solution to $\left(^{*}\right)$, which is identical to the original equation.

Answer: \frac{1}{2}


Mathematics -> Algebra -> Algebra -> Equations and Inequalities

Mathematics -> Discrete Mathematics -> Combinatorics

Find the maximum value of $m$ for a sequence $P_{0}, P_{1}, \cdots, P_{m+1}$ of points on a grid satisfying certain conditions.

Solution: We will show that the desired maximum value for $m$ is $n(n-1)$. First, let us show that $m \leq n(n-1)$ always holds for any sequence $P_{0}, P_{1}, \cdots, P_{m+1}$ satisfying the conditions of the problem. Call a point a turning point if it coincides with $P_{i}$ for some $i$ with $1 \leq i \leq m$. Let us say also that 2 points $\{P, Q\}$ are adjacent if $\{P, Q\}=\left\{P_{i-1}, P_{i}\right\}$ for some $i$ with $1 \leq i \leq m$, and vertically adjacent if, in addition, $P Q$ is parallel to the $y$-axis. Any turning point is vertically adjacent to exactly one other turning point. Therefore, the set of all turning points is partitioned into a set of pairs of points using the relation of "vertical adjacency". Thus we can conclude that if we fix $k \in\{1,2, \cdots, n\}$, the number of turning points having the $x$-coordinate $k$ must be even, and hence it is less than or equal to $n-1$. Therefore, altogether there are less than or equal to $n(n-1)$ turning points, and this shows that $m \leq n(n-1)$ must be satisfied. It remains now to show that for any positive odd number $n$ one can choose a sequence for which $m=n(n-1)$. We will show this by using the mathematical induction on $n$. For $n=1$, this is clear. For $n=3$, choose $P_{0}=(0,1), \quad P_{1}=(1,1), \quad P_{2}=(1,2), \quad P_{3}=(2,2), \quad P_{4}=(2,1), \quad P_{5}=(3,1), \quad P_{6}=(3,3), \quad P_{7}=(4,3)$. It is easy to see that these points satisfy the requirements. Let $n$ be an odd integer $\geq 5$, and suppose there exists a sequence satisfying the desired conditions for $n-4$. Then, it is possible to construct a sequence which gives a configuration indicated in the following diagram, where the configuration inside of the dotted square is given by the induction hypothesis: By the induction hypothesis, there are exactly $(n-4)(n-5)$ turning points for the configuration inside of the dotted square, and all of the lattice points lying outside of the dotted square except for the 4 points $(n, 2),(n-1, n-2),(2,3),(1, n-1)$ are turning points. Therefore, the total number of turning points in this configuration is $(n-4)(n-5)+\left(n^{2}-(n-4)^{2}-4\right)=n(n-1)$ showing that for this $n$ there exists a sequence satisfying the desired properties, and thus completing the induction process.

Answer: n(n-1)


Mathematics -> Algebra -> Linear Algebra -> Vectors

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 \(u\).

Solution: 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.

Answer: 2 \sqrt{3}


Mathematics -> Algebra -> Abstract Algebra -> Group Theory

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying $f(x) f(y)=f(x-y)$. Find all possible values of $f(2017)$.

Solution: Let $P(x, y)$ be the given assertion. From $P(0,0)$ we get $f(0)^{2}=f(0) \Longrightarrow f(0)=0,1$. From $P(x, x)$ we get $f(x)^{2}=f(0)$. Thus, if $f(0)=0$, we have $f(x)=0$ for all $x$, which satisfies the given constraints. Thus $f(2017)=0$ is one possibility. Now suppose $f(0)=1$. We then have $P(0, y) \Longrightarrow f(-y)=f(y)$, so that $P(x,-y) \Longrightarrow f(x) f(y)=$ $f(x-y)=f(x) f(-y)=f(x+y)$. Thus $f(x-y)=f(x+y)$, and in particular $f(0)=f\left(\frac{x}{2}-\frac{x}{2}\right)=$ $f\left(\frac{x}{2}+\frac{x}{2}\right)=f(x)$. It follows that $f(x)=1$ for all $x$, which also satisfies all given constraints. Thus the two possibilities are $f(2017)=0,1$.

Answer: 0, 1


The Houson Association of Mathematics Educators decides to hold a grand forum on mathematics education and invites a number of politicians from the United States to participate. Around lunch time the politicians decide to play a game. In this game, players can score 19 points for pegging the coordinator of the gathering with a spit ball, 9 points for downing an entire cup of the forum's interpretation of coffee, or 8 points for quoting more than three consecutive words from the speech Senator Bobbo delivered before lunch. What is the product of the two greatest scores that a player cannot score in this game?

Solution: Attainable scores are positive integers that can be written in the form \(8 a+9 b+19 c\), where \(a, b\), and \(c\) are nonnegative integers. Consider attainable number of points modulo 8. Scores that are \(0(\bmod 8)\) can be obtained with \(8 a\) for positive \(a\). Scores that are \(1(\bmod 8)\) greater than or equal to 9 can be obtained with \(9+8 a\) for nonnegative \(a\). Scores that are \(2(\bmod 8)\) greater than or equal to 18 can be obtained with \(9 \cdot 2+8 a\). Scores that are \(3(\bmod 8)\) greater than or equal to 19 can be obtained with \(19+8 a\). Scores that are \(4(\bmod 8)\) greater than or equal to \(19+9=28\) can be obtained with \(19+9+8 a\). Scores that are \(5(\bmod 8)\) greater than or equal to \(19+9 \cdot 2=37\) can be obtained with \(19+9 \cdot 2+8 a\). Scores that are \(6(\bmod 8)\) greater than or equal to \(19 \cdot 2=38\) can be obtained with \(19 \cdot 2+8 a\). Scores that are \(7(\bmod 8)\) greater than or equal to \(19 \cdot 2+9=47\) can be obtained with \(19 \cdot 2+9+8 a\). So the largest two unachievable values are 39 and 31. Multiplying them gives 1209.

Answer: 1209


The number 770 is written on a blackboard. Melody repeatedly performs moves, where a move consists of subtracting either 40 or 41 from the number on the board. She performs moves until the number is not positive, and then she stops. Let $N$ be the number of sequences of moves that Melody could perform. Suppose $N=a \cdot 2^{b}$ where $a$ is an odd positive integer and $b$ is a nonnegative integer. Compute $100 a+b$.

Solution: Notice that if we use the 41 move nine times or less, we will have to make a total of $\left\lceil\frac{770}{40}\right\rceil=20$ moves, and if we use it ten times or more, we will have to make a total of $\left\lfloor\frac{770}{40}\right\rfloor=19$ moves. So, doing casework on the number of 40 s we use gives $$\underbrace{\binom{19}{0}+\binom{19}{1}+\binom{19}{2}+\cdots+\binom{19}{9}}_{19 \text { moves }}+\underbrace{\frac{\binom{20}{10}}{2}+\binom{20}{11}+\binom{20}{11}+\cdots+\binom{20}{20}}_{20 \text { moves }}$$ Using the row sums of Pascal's triangle we have this sum equal to $\frac{2^{19}}{2}+\frac{2^{20}}{2}=3 \cdot 2^{18}$. The answer is 318.

Answer: 318


Mathematics -> Number Theory -> Factorization

How many perfect squares divide $2^{3} \cdot 3^{5} \cdot 5^{7} \cdot 7^{9}$?

Solution: The number of such perfect squares is $2 \cdot 3 \cdot 4 \cdot 5$, since the exponent of each prime can be any nonnegative even number less than the given exponent.

Answer: 120


Solution: 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.

Answer: -18910


Kevin starts with the vectors \((1,0)\) and \((0,1)\) and at each time step, he replaces one of the vectors with their sum. Find the cotangent of the minimum possible angle between the vectors after 8 time steps.

Solution: Say that the vectors Kevin has at some step are \((a, b)\) and \((c, d)\). Notice that regardless of which vector he replaces with \((a+c, b+d)\), the area of the triangle with vertices \((0,0),(a, b)\), and \((c, d)\) is preserved with the new coordinates. We can see this geometrically: the parallelogram with vertices \((0,0),(a, b)\), \((c, d)\), and \((a+c, b+d)\) can be cut in half by looking at the triangle formed by any 3 of the vertices, which include the original triangle, and both possible triangles that might arise in the next step. Because the area is preserved, the minimum possible angle then arises when the two vectors, our sides, are as long as possible. This occurs when we alternate which vector is getting replaced for the sum. Given two vectors \((a, b)\) and \((c, d)\), with \(\sqrt{a^{2}+b^{2}}>\sqrt{c^{2}+d^{2}}\), we would rather replace \((c, d)\) than \((a, b)\), and \((a+c, b+d)\) has a larger norm than \((a, b)\). Then at the \(n\)th step, Kevin has the vectors \(\left(F_{n}, F_{n-1}\right)\) and \(\left(F_{n+1}, F_{n}\right)\), where \(F_{0}=0\) and \(F_{1}=1\). The tangent of the angle between them is the tangent of the difference of the angles they make with the x-axis, which is just their slope. We can then compute the cotangent as \(\left|\frac{1+\frac{F_{n-1}}{F_{n}} \cdot \frac{F_{n}}{F_{n+1}}}{\frac{F_{n}}{F_{n+1}}-\frac{F_{n-1}}{F_{n}}}\right|=\left|\frac{F_{n}\left(F_{n+1}+F_{n-1}\right)}{F_{n}^{2}-F_{n-1} F_{n+1}}\right|\). We can show (by induction) that \(F_{n}\left(F_{n+1}+F_{n-1}\right)=F_{2 n}\) and \(F_{n}^{2}-F_{n-1} F_{n+1}=(-1)^{n+1}\). Thus at the 8th step, the cotangent of the angle is \(F_{16}=987\).

Answer: 987


Suppose \(\triangle A B C\) has lengths \(A B=5, B C=8\), and \(C A=7\), and let \(\omega\) be the circumcircle of \(\triangle A B C\). Let \(X\) be the second intersection of the external angle bisector of \(\angle B\) with \(\omega\), and let \(Y\) be the foot of the perpendicular from \(X\) to \(B C\). Find the length of \(Y C\).

Solution: Extend ray \(\overrightarrow{A B}\) to a point \(D\), since \(B X\) is an angle bisector, we have \(\angle X B C=\angle X B D=180^{\circ}-\angle X B A=\angle X C A\), so \(X C=X A\) by the inscribed angle theorem. Now, construct a point \(E\) on \(B C\) so that \(C E=A B\). Since \(\angle B A X \cong \angle B C X\), we have \(\triangle B A X \cong \triangle E C X\) by SAS congruence. Thus, \(X B=X E\), so \(Y\) bisects segment \(B E\). Since \(B E=B C-E C=8-5=3\), we have \(Y C=E C+Y E=5+\frac{1}{2} \cdot 3=\frac{13}{2}\). (Archimedes Broken Chord Theorem).

Answer: \frac{13}{2}


Mathematics -> Algebra -> Algebra -> Polynomial Operations

Mathematics -> Number Theory -> Congruences

Let \(\mathbb{Z}\) denote the set of all integers. Find all polynomials \(P(x)\) with integer coefficients that satisfy the following property: For any infinite sequence \(a_{1}, a_{2}, \ldots\) of integers in which each integer in \(\mathbb{Z}\) appears exactly once, there exist indices \(i<j\) and an integer \(k\) such that \(a_{i}+a_{i+1}+\cdots+a_{j}=P(k)\).

Solution: Part 1: All polynomials with \(\operatorname{deg} P=1\) satisfy the given property. Suppose \(P(x)=cx+d\), and assume without loss of generality that \(c>d \geq 0\). Denote \(s_{i}=a_{1}+a_{2}+\cdots+a_{i}(\bmod c)\). It suffices to show that there exist indices \(i\) and \(j\) such that \(j-i \geq 2\) and \(s_{j}-s_{i} \equiv d\) \((\bmod c)\). Consider \(c+1\) indices \(e_{1}, e_{2}, \ldots, e_{c+1}>1\) such that \(a_{e_{l}} \equiv d(\bmod c)\). By the pigeonhole principle, among the \(n+1\) pairs \((s_{e_{1}-1}, s_{e_{1}}), (s_{e_{2}-1}, s_{e_{2}}), \ldots, (s_{e_{n+1}-1}, s_{e_{n+1}})\), some two are equal, say \((s_{m-1}, s_{m})\) and \((s_{n-1}, s_{n})\). We can then take \(i=m-1\) and \(j=n\). Part 2: All polynomials with \(\operatorname{deg} P \neq 1\) do not satisfy the given property. Lemma: If \(\operatorname{deg} P \neq 1\), then for any positive integers \(A, B\), and \(C\), there exists an integer \(y\) with \(|y|>C\) such that no value in the range of \(P\) falls within the interval \([y-A, y+B]\). Proof of Lemma: The claim is immediate when \(P\) is constant or when \(\operatorname{deg} P\) is even since \(P\) is bounded from below. Let \(P(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0}\) be of odd degree greater than 1, and assume without loss of generality that \(a_{n}>0\). Since \(P(x+1)-P(x)=a_{n} n x^{n-1}+\ldots\), and \(n-1>0\), the gap between \(P(x)\) and \(P(x+1)\) grows arbitrarily for large \(x\). The claim follows. Suppose \(\operatorname{deg} P \neq 1\). We will inductively construct a sequence \(\{a_{i}\}\) such that for any indices \(i<j\) and any integer \(k\) it holds that \(a_{i}+a_{i+1}+\cdots+a_{j} \neq P(k)\). Suppose that we have constructed the sequence up to \(a_{i}\), and \(m\) is an integer with smallest magnitude yet to appear in the sequence. We will add two more terms to the sequence. Take \(a_{i+2}=m\). Consider all the new sums of at least two consecutive terms; each of them contains \(a_{i+1}\). Hence all such sums are in the interval \([a_{i+1}-A, a_{i+1}+B]\) for fixed constants \(A, B\). The lemma allows us to choose \(a_{i+1}\) so that all such sums avoid the range of \(P\). Alternate Solution for Part 1: Again, suppose \(P(x)=cx+d\), and assume without loss of generality that \(c>d \geq 0\). Let \(S_{i}=\{a_{j}+a_{j+1}+\cdots+a_{i}(\bmod c) \mid j=1,2, \ldots, i\}\). Then \(S_{i+1}=\{s_{i}+a_{i+1}(\bmod c) \mid s_{i} \in S_{i}\} \cup \{a_{i+1}(\bmod c)\}\). Hence \(|S_{i+1}|=|S_{i}|\) or \(S_{i+1}=|S_{i}|+1\), with the former occurring exactly when \(0 \in S_{i}\). Since \(|S_{i}| \leq c\), the latter can only occur finitely many times, so there exists \(I\) such that \(0 \in S_{i}\) for all \(i \geq I\). Let \(t>I\) be an index with \(a_{t} \equiv d(\bmod c)\). Then we can find a sum of at least two consecutive terms ending at \(a_{t}\) and congruent to \(d(\bmod c)\).

Answer: P(x) = cx + d \text{ with } c, d \in \mathbb{Z}


Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations

Mathematics -> Algebra -> Abstract Algebra -> Field Theory

Michael picks a random subset of the complex numbers \(\left\{1, \omega, \omega^{2}, \ldots, \omega^{2017}\right\}\) where \(\omega\) is a primitive \(2018^{\text {th }}\) root of unity and all subsets are equally likely to be chosen. If the sum of the elements in his subset is \(S\), what is the expected value of \(|S|^{2}\)? (The sum of the elements of the empty set is 0.)

Solution: Consider \(a\) and \(-a\) of the set of complex numbers. If \(x\) is the sum of some subset of the other complex numbers, then expected magnitude squared of the sum including \(a\) and \(-a\) is \(\frac{(x+a)(\overline{x+a})+x \bar{x}+x \bar{x}+(x-a)(\overline{x-a})}{4} = x \bar{x}+\frac{a \bar{a}}{2} = x \bar{x}+\frac{1}{2}\). By repeating this process on the remaining 2016 elements of the set, we can obtain a factor of \(\frac{1}{2}\) every time. In total, the answer is \(\frac{1009}{2}\).

Answer: \frac{1009}{2}


A polynomial $P$ has four roots, $\frac{1}{4}, \frac{1}{2}, 2,4$. The product of the roots is 1, and $P(1)=1$. Find $P(0)$.

Solution: A polynomial $Q$ with $n$ roots, $x_{1}, \ldots, x_{n}$, and $Q\left(x_{0}\right)=1$ is given by $Q(x)=\frac{\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right)}{\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right) \cdots\left(x_{0}-x_{4}\right)}$, so $P(0)=\frac{1}{\frac{3}{4} \cdot \frac{1}{2} \cdot(-1) \cdot(-3)}=\frac{8}{9}$.

Answer: \frac{8}{9}


Mathematics -> Geometry -> Solid Geometry -> 3D Shapes

In a wooden block shaped like a cube, all the vertices and edge midpoints are marked. The cube is cut along all possible planes that pass through at least four marked points. Let \(N\) be the number of pieces the cube is cut into. Estimate \(N\). An estimate of \(E>0\) earns \(\lfloor 20 \min (N / E, E / N)\rfloor\) points.

Solution: Answer: 15600

Answer: 15600


Let $a_{1}, a_{2}, \ldots$ be an arithmetic sequence and $b_{1}, b_{2}, \ldots$ be a geometric sequence. Suppose that $a_{1} b_{1}=20$, $a_{2} b_{2}=19$, and $a_{3} b_{3}=14$. Find the greatest possible value of $a_{4} b_{4}$.

Solution: Solution 1. Let $\{a_{n}\}$ have common difference $d$ and $\{b_{n}\}$ have common ratio $r$; for brevity, let $a_{1}=a$ and $b_{1}=b$. Then we have the equations $a b=20,(a+d) b r=19$, and $(a+2 d) b r^{2}=14$, and we want to maximize $(a+3 d) b r^{3}$. The equation $(a+d) b r=19$ expands as $a b r+d b r=19$, or $20 r+b d r=19$ since $a b=20$. Similarly, $(20+2 b d) r^{2}=14$, or $10 r^{2}+b d r^{2}=7$. Multiplying the first equation by $r$ and subtracting the second, we get $$10 r^{2}=19 r-7 \Longrightarrow(5 r-7)(2 r-1)=0$$ so either $r=\frac{7}{5}$ or $r=\frac{1}{2}$. For each value of $r$, we have $b d=\frac{19-20 r}{r}=\frac{19}{r}-20$, so $$(a+3 d) b r^{3}=(20+3 b d) r^{3}=\left(\frac{57}{r}-40\right) r^{3}=r^{2}(57-40 r)$$ The greater value of this expression is $\frac{37}{4}$, achieved when $r=\frac{1}{2}$.

Answer: \frac{37}{4}


A triple of integers \((a, b, c)\) satisfies \(a+b c=2017\) and \(b+c a=8\). Find all possible values of \(c\).

Solution: Add and subtract the two equations to find \((b+a)(c+1)=8+2017\) and \((b-a)(c-1)=2017-8\). We see that \(c\) is even and then that every integer \(c\) with \(c+1|2025, c-1| 2009\) works. We factor and solve. The full solutions are \((2017,8,0),(-667,1342,2),(-59,-346,-6),(-31,256,8)\).

Answer: -6,0,2,8


Mathematics -> Algebra -> Intermediate Algebra -> Inequalities

Find the largest positive integer \(n\) for which there exist \(n\) finite sets \(X_{1}, X_{2}, \ldots, X_{n}\) with the property that for every \(1 \leq a<b<c \leq n\), the equation \(\left|X_{a} \cup X_{b} \cup X_{c}\right|=\lceil\sqrt{a b c}\rceil\) holds.

Solution: First, we construct an example for \(N=4\). Let \(X_{1}, X_{2}, X_{3}, X_{4}\) be pairwise disjoint sets such that \(X_{1}=\varnothing,\left|X_{2}\right|=1,\left|X_{3}\right|=2\), and \(\left|X_{4}\right|=2\). It is straightforward to verify the condition. We claim that there are no five sets \(X_{1}, X_{2}, \ldots, X_{5}\) for which \(#\left(X_{a} \cup X_{b} \cup X_{c}\right)=\lceil\sqrt{a b c}\rceil\), for \(1 \leq a<b<c \leq 5\). Note that showing the non-existence of five such sets implies that there are no \(n\) sets with the desired property for \(n \geq 5\) as well. Suppose, for sake of contradiction, that there are such \(X_{1}, \ldots, X_{5}\). Then, note that \(\left|X_{1} \cup X_{2} \cup X_{4}\right|=3\), \(\left|X_{1} \cup X_{2} \cup X_{5}\right|=4\), and \(\left|X_{2} \cup X_{4} \cup X_{5}\right|=7\). Note that \(\left|X_{1} \cup X_{2} \cup X_{4}\right|+\left|X_{1} \cup X_{2} \cup X_{5}\right|=\left|X_{2} \cup X_{4} \cup X_{5}\right|\). For any sets \(A, B, C, D\), we have the following two inequalities: \(|A \cup B \cup C|+|A \cup B \cup D| \geq|A \cup B \cup C \cup D| \geq|B \cup C \cup D|\). For \(A=X_{1}, B=X_{2}, C=X_{4}\), and \(D=X_{5}\) in the situation above, we conclude that the equalities must both hold in both inequalities. The first equality shows that \(X_{1} \cup X_{2}=\varnothing\), and therefore both \(X_{1}\) and \(X_{2}\) are empty. Now observe that \(\left|X_{1} \cup X_{4} \cup X_{5}\right|=5 \neq 7=\left|X_{2} \cup X_{4} \cup X_{5}\right|\). This gives a contradiction.

Answer: 4


Mathematics -> Algebra -> Intermediate Algebra -> Other

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).

Solution: 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 $F_{11}=89$.

Answer: 89


Suppose there are 100 cookies arranged in a circle, and 53 of them are chocolate chip, with the remainder being oatmeal. Pearl wants to choose a contiguous subsegment of exactly 67 cookies and wants this subsegment to have exactly \(k\) chocolate chip cookies. Find the sum of the \(k\) for which Pearl is guaranteed to succeed regardless of how the cookies are arranged.

Solution: We claim that the only values of \(k\) are 35 and 36. WLOG assume that the cookies are labelled 0 through 99 around the circle. Consider the following arrangement: cookies 0 through 17,34 through 50, and 67 through 84 are chocolate chip, and the remaining are oatmeal. (The cookies form six alternating blocks around the circle of length \(18,16,17,16,18,15\).) Consider the block of 33 cookies that are not chosen. It is not difficult to see that since the sum of the lengths of each two adjacent block is always at least 33 and at most 34, this block of unchosen cookies always contains at least one complete block of cookies of the same type (and no other cookies of this type). So this block contains 17 or 18 or \(33-16=17\) or \(33-15=18\) chocolate chip cookies. Therefore, the block of 67 chosen cookies can only have \(53-17=36\) or \(53-18=35\) chocolate chip cookies. Now we show that 35 and 36 can always be obtained. Consider all possible ways to choose 67 cookies: cookies 0 through 66,1 through \(67, \ldots, 99\) through 65. It is not difficult to see that the number of chocolate chip cookies in the block changes by at most 1 as we advance from one way to the next. Moreover, each cookie will be chosen 67 times, so on average there will be \(\frac{67.53}{100}=35.51\) chocolate chip cookies in each block. Since not all blocks are below average and not all blocks are above average, there must be a point where a block below average transition into a block above average. The difference of these two blocks is at most 1, so one must be 35 and one must be 36. Therefore, the sum of all possible values of \(k\) is \(35+36=71\).

Answer: 71


Five people take a true-or-false test with five questions. Each person randomly guesses on every question. Given that, for each question, a majority of test-takers answered it correctly, let $p$ be the probability that every person answers exactly three questions correctly. Suppose that $p=\frac{a}{2^{b}}$ where $a$ is an odd positive integer and $b$ is a nonnegative integer. Compute 100a+b.

Solution: There are a total of $16^{5}$ ways for the people to collectively ace the test. Consider groups of people who share the same problems that they got incorrect. We either have a group of 2 and a group of 3 , or a group 5 . In the first case, we can pick the group of two in $\binom{5}{2}$ ways, the problems they got wrong in $\binom{5}{2}$ ways. Then there are 3! ways for the problems of group 3. There are 600 cases here. In the second case, we can $5!\cdot 4!/ 2=120 \cdot 12$ ways to organize the five cycle ( $4!/ 2$ to pick a cycle and 5 ! ways to assign a problem to each edge in the cycle). Thus, the solution is $\frac{255}{2^{17}}$ and the answer is 25517.

Answer: 25517


Mathematics -> Discrete Mathematics -> Graph Theory

How many graphs are there on 10 vertices labeled \(1,2, \ldots, 10\) such that there are exactly 23 edges and no triangles?

Solution: Note that the sum of the degrees of the graph is \(23 \cdot 2=46\), so at least one vertex has degree 5 or more. We casework on the maximal degree \(n\). Case 1: \(n \geq 7\), then none of the \(n\) neighbors can have an edge between each other, for \(\binom{n}{2}\) edges unusable, and the vertex with maximal degree cannot connect to the \(9-n\) other vertices. Then we have \(\binom{n}{2}+9-n>\binom{10}{2}-23=22\) when \(n \geq 7\), so there cannot be any graph in this case. Case 2: \(n=6\). WLOG suppose that 1 is connected to \(2,3,4,5,6,7\), then none of \(2,3,4,5,6,7\) can connect to each other. Case 2.1: There is at least one edge between \(8,9,10\), then each of \(2,3,4,5,6,7\) can connect to at most two of \(8,9,10\), for at most \(6 \cdot 2+\binom{3}{2}=15\) additional edges. Along with the 6 original edges, it is not enough to each 23 edges. Case 2.2: There are no edges between \(8,9,10\), then this graph is a bipartite graph between \(1,8,9,10\) and \(2,3,4,5,6,7\). There can be at most \(4 \cdot 6=24\) edges in this graph, so exactly one edge is removed from this graph. There are \(\binom{10}{4} \cdot 24=5040\) possible graphs in this case. Case 3: \(n=5\). WLOG suppose that 1 is connected to \(2,3,4,5,6\), then none of \(2,3,4,5,6\) can connect to each other. Case 3.1: There is at least one edge between \(7,8,9,10\). Then each of \(2,3,4,5,6\) can connect to at most three of \(7,8,9,10\), for \(5 \cdot 3=15\) edges. In this case at least three of \(7,8,9,10\) must not be connected to each other, so there can be at most three edges, for \(5+15+3=23\) edges at most. However, this requires the three disconnected vertices of \(7,8,9,10\) to be connected to all of \(2,3,4,5,6\) and the other vertex of \(7,8,9,10\), causing them to have degree 6. We can therefore ignore this case. (The case where \(2,3,4,5,6\) can connect to two or less of \(7,8,9,10\) can be easily ruled out.) Case 3.2: There are no edges between \(7,8,9,10\), then this graph is a bipartite graph between \(1,7,8,9,10\) and \(2,3,4,5,6\). This is a \(K_{5,5}\) with two edges removed, which accounts for \(\binom{10}{5} / 2 \cdot\binom{25}{2}=126 \cdot 300=37800\) graphs. It is not difficult to see that Case 2.2 and Case 3.2 are disjoint (by considering max degree), so there are \(5040+37800=42840\) graphs in total.

Answer: 42840


Mathematics -> Number Theory -> Diophantine Equations -> Other

Find all positive integers $a$ and $b$ such that $\frac{a^{2}+b}{b^{2}-a}$ and $\frac{b^{2}+a}{a^{2}-b}$ are both integers.

Solution: By the symmetry of the problem, we may suppose that $a \leq b$. Notice that $b^{2}-a \geq 0$, so that if $\frac{a^{2}+b}{b^{2}-a}$ is a positive integer, then $a^{2}+b \geq b^{2}-a$. Rearranging this inequality and factorizing, we find that $(a+b)(a-b+1) \geq 0$. Since $a, b>0$, we must have $a \geq b-1$. We therefore have two cases: Case 1: $a=b$. Substituting, we have $\frac{a^{2}+a}{a^{2}-a}=\frac{a+1}{a-1}=1+\frac{2}{a-1}$ which is an integer if and only if $(a-1) \mid 2$. As $a>0$, the only possible values are $a-1=1$ or 2. Hence, $(a, b)=(2,2)$ or $(3,3)$. Case 2: $a=b-1$. Substituting, we have $\frac{b^{2}+a}{a^{2}-b}=\frac{(a+1)^{2}+a}{a^{2}-(a+1)}=\frac{a^{2}+3 a+1}{a^{2}-a-1}=1+\frac{4 a+2}{a^{2}-a-1}$. For $\frac{4 a+2}{a^{2}-a-1}$ to be an integer, we must have $4 a+2 \geq a^{2}-a-1$, that is, $a^{2}-5 a-3 \leq 0$. Hence, since $a$ is an integer, we can bound $a$ by $1 \leq a \leq 5$. Checking all the ordered pairs $(a, b)=(1,2),(2,3), \ldots,(5,6)$, we find that only $(1,2)$ and $(2,3)$ satisfy the given conditions. Thus, the ordered pairs that work are $(2,2),(3,3),(1,2),(2,3),(2,1),(3,2)$.

Answer: (2,2),(3,3),(1,2),(2,3),(2,1),(3,2)


Mathematics -> Algebra -> Prealgebra -> Integers

Find the sum of the digits of \(11 \cdot 101 \cdot 111 \cdot 110011\).

Solution: There is no regrouping, so the answer is \(2 \cdot 2 \cdot 3 \cdot 4=48\). The actual product is 13566666531.

Answer: 48


Mathematics -> Geometry -> Plane Geometry -> Angles

Mathematics -> Geometry -> Plane Geometry -> Triangulations

Triangle \(\triangle A B C\) has \(A B=21, B C=55\), and \(C A=56\). There are two points \(P\) in the plane of \(\triangle A B C\) for which \(\angle B A P=\angle C A P\) and \(\angle B P C=90^{\circ}\). Find the distance between them.

Solution: Let \(P_{1}\) and \(P_{2}\) be the two possible points \(P\), with \(A P_{1}<A P_{2}\). Both lie on the \(\angle A\)-bisector and the circle \(\gamma\) with diameter \(B C\). Let \(D\) be the point where the \(\angle A\)-bisector intersects \(B C\), let \(M\) be the midpoint of \(B C\), and let \(X\) be the foot of the perpendicular from \(M\) onto the \(\angle A\)-bisector. Since we know the radius of \(\gamma\), to compute \(P_{1} P_{2}\) it suffices to compute \(M X\). By the angle bisector theorem we find \(B D=15\) and \(D C=40\), so Stewart's theorem gives \(15 \cdot 40 \cdot 55+55 \cdot A D^{2}=21^{2} \cdot 40+56^{2} \cdot 15 \Longrightarrow A D=24\). Then \(\cos \angle A D B=\frac{-21^{2}+15^{2}+24^{2}}{2 \cdot 15 \cdot 24}=\frac{1}{2}\), so \(\angle A D B=\angle M D X=60^{\circ}\). Since \(D M=B M-B D=\frac{55}{2}-15=\frac{25}{2}\), we get \(M X=D M \sin \angle M D X=\frac{25 \sqrt{3}}{4}\). Hence \(P_{1} P_{2}=2 \sqrt{\left(\frac{55}{2}\right)^{2}-\left(\frac{25 \sqrt{3}}{4}\right)^{2}}=\frac{5 \sqrt{409}}{2}\).

Answer: \frac{5}{2} \sqrt{409}


Mathematics -> Algebra -> Algebra -> Algebraic Expressions

Mathematics -> Calculus -> Integral Calculus -> Techniques of Integration -> Single-variable

Find the value of $$\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{a b(3 a+c)}{4^{a+b+c}(a+b)(b+c)(c+a)}$$

Solution: Let $S$ denote the given sum. By summing over all six permutations of the variables $a, b, c$ we obtain $$\begin{aligned} 6 S & =\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{3\left(a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b\right)+6 a b c}{4^{a+b+c}(a+b)(b+c)(c+a)} \\ & =\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{3}{4^{a+b+c}} \\ & =3\left(\sum_{a=1}^{\infty} \frac{1}{4^{a}}\right)\left(\sum_{b=1}^{\infty} \frac{1}{4^{b}}\right)\left(\sum_{c=1}^{\infty} \frac{1}{4^{c}}\right) \\ & =3\left(\frac{1}{3}\right)^{3} \\ & =\frac{1}{9} \end{aligned}$$ Hence $S=\frac{1}{54}$.

Answer: \frac{1}{54}


Allen and Yang want to share the numbers \(1,2,3,4,5,6,7,8,9,10\). How many ways are there to split all ten numbers among Allen and Yang so that each person gets at least one number, and either Allen's numbers or Yang's numbers sum to an even number?

Solution: Since the sum of all of the numbers is odd, exactly one of Allen's sum and Yang's sum must be odd. Therefore any way of splitting the numbers up where each person receives at least one number is valid, so the answer is \(2^{10}-2=1022\).

Answer: 1022


Mathematics -> Discrete Mathematics -> Combinatorics

Find the total number of different integer values the function $$f(x)=[x]+[2 x]+\left[\frac{5 x}{3}\right]+[3 x]+[4 x]$$ takes for real numbers $x$ with $0 \leq x \leq 100$. Note: $[t]$ is the largest integer that does not exceed $t$.

Solution: Note that, since $[x+n]=[x]+n$ for any integer $n$, $$f(x+3)=[x+3]+[2(x+3)]+\left[\frac{5(x+3)}{3}\right]+[3(x+3)]+[4(x+3)]=f(x)+35$$ one only needs to investigate the interval $[0,3)$. The numbers in this interval at which at least one of the real numbers $x, 2 x, \frac{5 x}{3}, 3 x, 4 x$ is an integer are - $0,1,2$ for $x$; - $\frac{n}{2}, 0 \leq n \leq 5$ for $2 x$; - $\frac{3 n}{5}, 0 \leq n \leq 4$ for $\frac{5 x}{3}$; - $\frac{n}{3}, 0 \leq n \leq 8$ for $3 x$; - $\frac{n}{4}, 0 \leq n \leq 11$ for $4 x$. Of these numbers there are - 3 integers $(0,1,2)$; - 3 irreducible fractions with 2 as denominator (the numerators are $1,3,5$ ); - 6 irreducible fractions with 3 as denominator (the numerators are 1, 2, 4, 5, 7, 8); - 6 irreducible fractions with 4 as denominator (the numerators are $1,3,5,7,9,11,13,15$ ); - 4 irreducible fractions with 5 as denominator (the numerators are 3, 6, 9, 12). Therefore $f(x)$ increases 22 times per interval. Since $100=33 \cdot 3+1$, there are $33 \cdot 22$ changes of value in $[0,99)$. Finally, there are 8 more changes in [99,100]: 99, 100, $99 \frac{1}{2}, 99 \frac{1}{3}, 99 \frac{2}{3}, 99 \frac{1}{4}$, $99 \frac{3}{4}, 99 \frac{3}{5}$. The total is then $33 \cdot 22+8=734$.

Answer: 734


Let $\omega_{1}, \omega_{2}, \ldots, \omega_{100}$ be the roots of $\frac{x^{101}-1}{x-1}$ (in some order). Consider the set $S=\left\{\omega_{1}^{1}, \omega_{2}^{2}, \omega_{3}^{3}, \ldots, \omega_{100}^{100}\right\}$. Let $M$ be the maximum possible number of unique values in $S$, and let $N$ be the minimum possible number of unique values in $S$. Find $M-N$.

Solution: Throughout this solution, assume we're working modulo 101. First, $N=1$. Let $\omega$ be a primitive 101 st root of unity. We then let $\omega_{n}=\omega^{1 / n}$, which we can do because 101 is prime, so $1 / n$ exists for all nonzero $n$ and $1 / n=1 / m \Longrightarrow m=n$. Thus the set contains only one distinct element, $\omega$. $M=100$ is impossible. Fix $\zeta$, a primitive 101st root of unity, and let $\omega_{n}=\zeta^{\pi(n)}$ for each $n$. Suppose that there are 100 distinct such $n \pi(n)$ exponents; then $\pi$ permutes the set $\{1,2, \cdots, 100\}$. Fix $g$, a primitive root of 101 ; write $n=g^{e_{n}}$ and $\pi(n)=g^{\tau\left(e_{n}\right)}$. Then $\left\{e_{n}\right\}=\{0,1,2, \ldots, 100\}$ and $\tau$ is a permutation of this set, as is $e_{n}+\tau\left(e_{n}\right)$. However, this is impossible: $\sum_{n=1}^{100} e_{n}+\tau\left(e_{n}\right)=5050+5050 \equiv$ $5050(\bmod 100)$, which is a contradiction. Thus there cannot be 100 distinct exponents. $M=99$ is possible. Again, let $\zeta$ be a primitive root of unity and let $\omega_{n}=\zeta^{1 /(n+1)}$, except when $n=100$, in which case let $\omega_{100}$ be the last possible root. Notice that $\frac{n}{n+1}=\frac{m}{m+1}$ if and only if $n=m$, so this will produce 99 different elements in the set. Thus $M-N=99-1=98$.

Answer: 98


Mathematics -> Precalculus -> Trigonometric Functions

Let $P A B C$ be a tetrahedron such that $\angle A P B=\angle A P C=\angle B P C=90^{\circ}, \angle A B C=30^{\circ}$, and $A P^{2}$ equals the area of triangle $A B C$. Compute $\tan \angle A C B$.

Solution: Observe that $$\begin{aligned} \frac{1}{2} \cdot A B \cdot A C \cdot \sin \angle B A C & =[A B C]=A P^{2} \\ & =\frac{1}{2}\left(A B^{2}+A C^{2}-B C^{2}\right) \\ & =A B \cdot A C \cdot \cos \angle B A C \end{aligned}$$ so $\tan \angle B A C=2$. Also, we have $\tan \angle A B C=\frac{1}{\sqrt{3}}$. Also, for any angles $\alpha, \beta, \gamma$ summing to $180^{\circ}$, one can see that $\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \cdot \tan \beta \cdot \tan \gamma$. Thus we have $\tan \angle A C B+2+\frac{1}{\sqrt{3}}=$ $\tan \angle A C B \cdot 2 \cdot \frac{1}{\sqrt{3}}$, so $\tan \angle A C B=8+5 \sqrt{3}$.

Answer: 8+5 \sqrt{3}


Determine all positive integers \(k\) for which there exist a positive integer \(m\) and a set \(S\) of positive integers such that any integer \(n>m\) can be written as a sum of distinct elements of \(S\) in exactly \(k\) ways.

Solution: We claim that \(k=2^{a}\) for all \(a \geq 0\). Let \(A=\{1,2,4,8, \ldots\}\) and \(B=\mathbb{N} \backslash A\). For any set \(T\), let \(s(T)\) denote the sum of the elements of \(T\). (If \(T\) is empty, we let \(s(T)=0\).) We first show that any positive integer \(k=2^{a}\) satisfies the desired property. Let \(B^{\prime}\) be a subset of \(B\) with \(a\) elements, and let \(S=A \cup B^{\prime}\). Recall that any nonnegative integer has a unique binary representation. Hence, for any integer \(t>s\left(B^{\prime}\right)\) and any subset \(B^{\prime \prime} \subseteq B^{\prime}\), the number \(t-s\left(B^{\prime \prime}\right)\) can be written as a sum of distinct elements of \(A\) in a unique way. This means that \(t\) can be written as a sum of distinct elements of \(B^{\prime}\) in exactly \(2^{a}\) ways. Next, assume that some positive integer \(k\) satisfies the desired property for a positive integer \(m \geq 2\) and a set \(S\). Clearly, \(S\) is infinite. Lemma: For all sufficiently large \(x \in S\), the smallest element of \(S\) larger than \(x\) is \(2x\). Proof of Lemma: Let \(x \in S\) with \(x>3m\), and let \(x<y<2x\). We will show that \(y \notin S\). Suppose first that \(y>x+m\). Then \(y-x\) can be written as a sum of distinct elements of \(S\) not including \(x\) in \(k\) ways. If \(y \in S\), then \(y\) can be written as a sum of distinct elements of \(S\) in at least \(k+1\) ways, a contradiction. Suppose now that \(y \leq x+m\). We consider \(z \in(2x-m, 2x)\). Similarly as before, \(z-x\) can be written as a sum of distinct elements of \(S\) not including \(x\) or \(y\) in \(k\) ways. If \(y \in S\), then since \(m<z-y<x\), \(z-y\) can be written as a sum of distinct elements of \(S\) not including \(x\) or \(y\). This means that \(z\) can be written as a sum of distinct elements of \(S\) in at least \(k+1\) ways, a contradiction. We now show that \(2x \in S\); assume for contradiction that this is not the case. Observe that \(2x\) can be written as a sum of distinct elements of \(S\) including \(x\) in exactly \(k-1\) ways. This means that \(2x\) can also be written as a sum of distinct elements of \(S\) not including \(x\). If this sum includes any number less than \(x-m\), then removing this number, we can write some number \(y \in(x+m, 2x)\) as a sum of distinct elements of \(S\) not including \(x\). Now if \(y=y^{\prime}+x\) where \(y^{\prime} \in(m, x)\) then \(y^{\prime}\) can be written as a sum of distinct elements of \(S\) including \(x\) in exactly \(k\) ways. Therefore \(y\) can be written as a sum of distinct elements of \(S\) in at least \(k+1\) ways, a contradiction. Hence the sum only includes numbers in the range \([x-m, x)\). Clearly two numbers do not suffice. On the other hand, three such numbers sum to at least \(3(x-m)>2x\), a contradiction. From the Lemma, we have that \(S=T \cup U\), where \(T\) is finite and \(U=\{x, 2x, 4x, 8x, \ldots\}\) for some positive integer \(x\). Let \(y\) be any positive integer greater than \(s(T)\). For any subset \(T^{\prime} \subseteq T\), if \(y-s\left(T^{\prime}\right) \equiv 0(\bmod x)\), then \(y-s\left(T^{\prime}\right)\) can be written as a sum of distinct elements of \(U\) in a unique way; otherwise \(y-s\left(T^{\prime}\right)\) cannot be written as a sum of distinct elements of \(U\). Hence the number of ways to write \(y\) as a sum of distinct elements of \(S\) is equal to the number of subsets \(T^{\prime} \subseteq T\) such that \(s\left(T^{\prime}\right) \equiv y(\bmod x)\). Since this holds for all \(y\), for any \(0 \leq a \leq x-1\) there are exactly \(k\) subsets \(T^{\prime} \subseteq T\) such that \(s\left(T^{\prime}\right) \equiv a(\bmod x)\). This means that there are \(kx\) subsets of \(T\) in total. But the number of subsets of \(T\) is a power of 2, and therefore \(k\) is a power of 2, as claimed.

Answer: k = 2^{a} \text{ for all } a \geq 0


Let $a_{0}, a_{1}, a_{2}, \ldots$ be an infinite sequence where each term is independently and uniformly random in the set $\{1,2,3,4\}$. Define an infinite sequence $b_{0}, b_{1}, b_{2}, \ldots$ recursively by $b_{0}=1$ and $b_{i+1}=a_{i}^{b_{i}}$. Compute the expected value of the smallest positive integer $k$ such that $b_{k} \equiv 1(\bmod 5)$.

Solution: Do casework on what $a_{0}$ is. If $a_{0}=1$ then $k=1$. If $a_{0}=4$ then $k=2$. If $a_{0}=3$ then - if $a_{1}=1$, then $k=2$ - if $a_{1}=2$ or 4 , then $k=3$ - if $a_{1}=3$, then you make no progress. so in expectation it requires $E=(2+3+(E+1)+3) / 4 \Longrightarrow E=3$. If $a_{0}=2$ then - if $a_{1}=1$ or 4 , then $k=2$ - if $a_{1}=2$, then $k=3$ - if $a_{1}=3$, then it can be checked that if $a_{2}=1$ we get $k=3$, if $a_{2}=2$ or 4 then $k=4$, and if $a_{2}=3$ then we make no progress. Thus, this case is equivalent to the case of $a_{0}=3$ except shifted over by one, so it is $3+1=4$ in expectation. So this case is $(2+3+4+2) / 4$ in expectation. This means the answer is $(1+(11 / 4)+3+2) / 4=35 / 16$.

Answer: \frac{35}{16}


Mathematics -> Applied Mathematics -> Statistics -> Mathematical Statistics

679 contestants participated in HMMT February 2017. Let \(N\) be the number of these contestants who performed at or above the median score in at least one of the three individual tests. Estimate \(N\). An estimate of \(E\) earns \(\left\lfloor 20-\frac{|E-N|}{2}\right\rfloor\) or 0 points, whichever is greater.

Solution: Out of the 679 total contestants at HMMT February 2017, 188 contestants scored at least the median on all three tests, 159 contestants scored at least the median on two tests, and 169 contestants scored at least the median on one test, giving a total of 516 contestants

Answer: 516


Mathematics -> Algebra -> Algebra -> Equations and Inequalities

Determine all positive integers $n$ for which the equation $$x^{n}+(2+x)^{n}+(2-x)^{n}=0$$ has an integer as a solution.

Solution: If $n$ is even, $x^{n}+(2+x)^{n}+(2-x)^{n}>0$, so $n$ is odd. For $n=1$, the equation reduces to $x+(2+x)+(2-x)=0$, which has the unique solution $x=-4$. For $n>1$, notice that $x$ is even, because $x, 2-x$, and $2+x$ have all the same parity. Let $x=2 y$, so the equation reduces to $$y^{n}+(1+y)^{n}+(1-y)^{n}=0$$ Looking at this equation modulo 2 yields that $y+(1+y)+(1-y)=y+2$ is even, so $y$ is even. Using the factorization $$a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2} b+\cdots+b^{n-1}\right) \text { for } n \text { odd }$$ which has a sum of $n$ terms as the second factor, the equation is now equivalent to $$y^{n}+(1+y+1-y)\left((1+y)^{n-1}-(1+y)^{n-2}(1-y)+\cdots+(1-y)^{n-1}\right)=0$$ or $$y^{n}=-2\left((1+y)^{n-1}-(1+y)^{n-2}(1-y)+\cdots+(1-y)^{n-1}\right)$$ Each of the $n$ terms in the second factor is odd, and $n$ is odd, so the second factor is odd. Therefore, $y^{n}$ has only one factor 2 , which is a contradiction to the fact that, $y$ being even, $y^{n}$ has at least $n>1$ factors 2 . Hence there are no solutions if $n>1$.

Answer: n=1


Mathematics -> Algebra -> Intermediate Algebra -> Logarithmic Functions

How many positive integers $n \leq 2009$ have the property that $\left\lfloor\log _{2}(n)\right\rfloor$ is odd?

Solution: We wish to find $n$ such that there is some natural number $k$ for which $2 k-1 \leq \log _{2} n<$ $2 k$. Since $n \leq 2009$ we must have $k \leq 5$. This is equivalent to finding the number of positive integers $n \leq 2009$ satisfying $2^{2 k-1} \leq n<2^{2 k}$ for some $k \leq 5$, so the number of such integers is $2+2^{3}+2^{5}+2^{7}+2^{9}=682$

Answer: 682


Mathematics -> Geometry -> Plane Geometry -> Polygons

Let $A X B Y$ be a cyclic quadrilateral, and let line $A B$ and line $X Y$ intersect at $C$. Suppose $A X \cdot A Y=6, B X \cdot B Y=5$, and $C X \cdot C Y=4$. Compute $A B^{2}$.

Solution: Observe that $$\begin{aligned} & \triangle A C X \sim \triangle Y C B \Longrightarrow \frac{A C}{A X}=\frac{C Y}{B Y} \\ & \triangle A C Y \sim \triangle X C B \Longrightarrow \frac{A C}{A Y}=\frac{C X}{B X} \end{aligned}$$ Mulitplying these two equations together, we get that $$A C^{2}=\frac{(C X \cdot C Y)(A X \cdot A Y)}{B X \cdot B Y}=\frac{24}{5}$$ Analogously, we obtain that $$B C^{2}=\frac{(C X \cdot C Y)(B X \cdot B Y)}{A X \cdot A Y}=\frac{10}{3}$$ Hence, we have $$A B=A C+B C=\sqrt{\frac{24}{5}}+\sqrt{\frac{10}{3}}=\frac{11 \sqrt{30}}{15}$$ implying the answer.

Answer: \frac{242}{15}


Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other

There are 2018 frogs in a pool and there is 1 frog on the shore. In each time-step thereafter, one random frog moves position. If it was in the pool, it jumps to the shore, and vice versa. Find the expected number of time-steps before all frogs are in the pool for the first time.

Solution: Consider the general case of \(n\) frogs. Let \(E_{i}\) be the expected time for all frogs to enter the pool when \(i\) frogs are on the shore and \(n-i\) frogs are in the pool. We have \(E_{0}=0, E_{n}=1+E_{n-1}\), and \(E_{i}=\frac{i}{n} E_{i-1}+\frac{n-i}{n} E_{i+1}+1\) for \(0<i<n\). Define \(f_{i}\) so that \(E_{i}=\frac{f_{i}}{(n-1)(n-2) \cdots(i)}+E_{i-1}\). Then by plugging this equation into the first equation, we can show that \(f_{i}=n(n-1) \cdots(i+1)+(n-i) f_{i+1}\). Furthermore, we know that \(f_{n}=1\). Therefore \(f_{1} =\sum_{i=1}^{n} \frac{n!}{i!} \frac{(n-1)!}{(n-i)!} =(n-1)!\sum_{i=1}^{n}\binom{n}{i} =(n-1)!\left(2^{n}-1\right)\). Therefore \(E_{1}=\frac{(n-1)!\left(2^{n}-1\right)}{(n-1)!}+E_{0}=2^{n}-1\). Plugging in \(n=2018\) yields \(E_{1}=2^{2018}-1\).

Answer: 2^{2018}-1