Let $m_1,m_2,...,m_{2013} > 1$ be 2013 pairwise relatively prime positive integers and $A_1,A_2,...,A_{2013}$ be 2013 (possibly empty) sets with $A_i\subseteq \{1,2,...,m_i-1\}$ for $i=1,2,...,2013$. Prove that there is a positive integer $N$ such that \[ N \le \left( 2\left\lvert A_1 \right\rvert + 1 \right)\left( 2\left\lvert A_2 \right\rvert + 1 \right)\cdots\left( 2\left\lvert A_{2013} \right\rvert + 1 \right) \] and for each $i = 1, 2, ..., 2013$, there does [i]not[/i] exist $a \in A_i$ such that $m_i$ divides $N-a$. [i]Proposed by Victor Wang[/i]

Let an even positive integer $2k$ be given. Find such relatively prime positive integers $x, y$ that maximize the product $xy$.

Given a polynomial $f(x)$ with rational coefficients, of degree $d \ge 2$, we define the sequence of sets $f^0(\mathbb{Q}), f^1(\mathbb{Q}), \ldots$ as $f^0(\mathbb{Q})=\mathbb{Q}$, $f^{n+1}(\mathbb{Q})=f(f^{n}(\mathbb{Q}))$ for $n\ge 0$. (Given a set $S$, we write $f(S)$ for the set $\{f(x)\mid x\in S\})$. Let $f^{\omega}(\mathbb{Q})=\bigcap_{n=0}^{\infty} f^n(\mathbb{Q})$ be the set of numbers that are in all of the sets $f^n(\mathbb{Q})$, $n\geq 0$. Prove that $f^{\omega}(\mathbb{Q})$ is a finite set. [i]Dan Schwarz, Romania[/i]

When $4 \cos \theta - 3 \sin \theta = \tfrac{13}{3},$ it follows that $7 \cos 2\theta - 24 \sin 2\theta = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Let $a/b$ be the probability that a randomly selected divisor of $2007$ is a multiple of $3$. If $a$ and $b$ are relatively prime positive integers, find $a+b$.

Find all pairs of positive integers $(x,y)$ with the following property: If $a,b$ are relative prime and positive divisors of $ x^3 + y^3$, then $a+b - 1$ is divisor of $x^3+y^3$. (Cyprus)

Find all pairs $(x, y)$ of positive rational numbers such that $x^{y}=y^{x}$.

A snowman is built on a level plane by placing a ball radius $6$ on top of a ball radius $8$ on top of a ball radius $10$ as shown. If the average height above the plane of a point in the snowman is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, find $m + n$. [asy] size(150); draw(circle((0,0),24)); draw(ellipse((0,0),24,9)); draw(circle((0,-56),32)); draw(ellipse((0,-56),32,12)); draw(circle((0,-128),40)); draw(ellipse((0,-128),40,15)); [/asy]

Wendy takes Honors Biology at school, a smallish class with only fourteen students (including Wendy) who sit around a circular table. Wendy's friends Lucy, Starling, and Erin are also in that class. Last Monday none of the fourteen students were absent from class. Before the teacher arrived, Lucy and Starling stretched out a blue piece of yarn between them. Then Wendy and Erin stretched out a red piece of yarn between them at about the same height so that the yarn would intersect if possible. If all possible positions of the students around the table are equally likely, let $m/n$ be the probability that the yarns intersect, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Let rational numbers $a, b, c$ satisfy the conditions $a + b + c = a^2 + b^2 + c^2 \in Z$. Prove that there exist two relative prime numbers $m, n$ such that $abc =\frac{m^2}{n^3}$ .

$A,B,C$ are points in the plane such that $\angle ABC=90^\circ$. Circles with diameters $BA$ and $BC$ meet at $D$. If $BA=20$ and $BC=21$, then the length of segment $BD$ can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$? [i]Ray Li[/i]

In triangle $ ABC$, angle $ C$ is a right angle and $ CB > CA$. Point $ D$ is located on $ \overline{BC}$ so that angle $ CAD$ is twice angle $ DAB$. If $ AC/AD \equal{} 2/3$, then $ CD/BD \equal{} m/n$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m \plus{} n$. $ \textbf{(A)}\ 10\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 22\qquad \textbf{(E)}\ 26$

When the number $2^{1000}$ is divided by $13$, the remainder in the division is $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }7\qquad \textbf{(E) }11$

If $\phi$ is the Golden Ratio, we know that $\frac1\phi = \phi - 1$. Define a new positive real number, called $\phi_d$, where $\frac1{\phi_d} = \phi_d - d$ (so $\phi = \phi_1$). Given that $\phi_{2009} = \frac{a + \sqrt{b}}{c}$, $a, b, c$ positive integers, and the greatest common divisor of $a$ and $c$ is 1, find $a + b + c$.

You have four fair $6$-sided dice, each numbered $1$ to $6$ (inclusive). If all four dice are rolled, the probability that the product of the rolled numbers is prime can be written as $\tfrac{a}{b}$, where $a$ and $b$ are relatively prime. What is $a+b$?

Find the number of positive integers $n$ satisfying $\phi(n) | n$ such that \[\sum_{m=1}^{\infty} \left( \left[ \frac nm \right] - \left[\frac{n-1}{m} \right] \right) = 1992\] What is the largest number among them? As usual, $\phi(n)$ is the number of positive integers less than or equal to $n$ and relatively prime to $n.$

Let $\phi(n)$ be the number of positive integers less than $n$ that are relatively prime to $n$, where $n$ is a positive integer. Find all pairs of positive integers $(m,n)$ such that \[2^n + (n-\phi(n)-1)! = n^m+1.\]

Let $ N$ be a positive integer. How many non-negative integers $ n \le N$ are there that have an integer multiple, that only uses the digits $ 2$ and $ 6$ in decimal representation?

Find all positive integers $n, n>1$ for wich holds : If $a_1, a_2 ,\dots ,a_k$ are all numbers less than $n$ and relatively prime to $n$ , and holds $a_1<a_2<\dots <a_k $, then none of sums $a_i+a_{i+1}$ for $i=1,2,3,\dots k-1 $ are divisible by $3$.

Two rational numbers \(\tfrac{m}{n}\) and \(\tfrac{n}{m}\) are written on a blackboard, where \(m\) and \(n\) are relatively prime positive integers. At any point, Evan may pick two of the numbers \(x\) and \(y\) written on the board and write either their arithmetic mean \(\tfrac{x+y}{2}\) or their harmonic mean \(\tfrac{2xy}{x+y}\) on the board as well. Find all pairs \((m,n)\) such that Evan can write 1 on the board in finitely many steps. [i]Proposed by Yannick Yao[/i]

Let $ABCD$ be a convex quadrilateral with $\angle ABC = 120^{\circ}$ and $\angle BCD = 90^{\circ}$, and let $M$ and $N$ denote the midpoints of $\overline{BC}$ and $\overline{CD}$. Suppose there exists a point $P$ on the circumcircle of $\triangle CMN$ such that ray $MP$ bisects $\overline{AD}$ and ray $NP$ bisects $\overline{AB}$. If $AB + BC = 444$, $CD = 256$ and $BC = \frac mn$ for some relatively prime positive integers $m$ and $n$, compute $100m+n$. [i]Proposed by Michael Ren[/i]

Find all odd positive integers $ n > 1$ such that if $ a$ and $ b$ are relatively prime divisors of $ n$, then $ a\plus{}b\minus{}1$ divides $ n$.

Joe the teacher is bad at rounding. Because of this, he has come up with his own way to round grades, where a [i]grade[/i] is a nonnegative decimal number with finitely many digits after the decimal point. Given a grade with digits $a_1a_2 \dots a_m.b_1b_2 \dots b_n$, Joe first rounds the number to the nearest $10^{-n+1}$th place. He then repeats the procedure on the new number, rounding to the nearest $10^{-n+2}$th, then rounding the result to the nearest $10^{-n+3}$th, and so on, until he obtains an integer. For example, he rounds the number $2014.456$ via $2014.456 \to 2014.46 \to 2014.5 \to 2015$. There exists a rational number $M$ such that a grade $x$ gets rounded to at least $90$ if and only if $x \ge M$. If $M = \tfrac pq$ for relatively prime integers $p$ and $q$, compute $p+q$. [i]Proposed by Yang Liu[/i]

Suppose that $m$ and $n$ are relatively prime positive integers with $A = \tfrac mn$, where \[ A = \frac{2+4+6+\dots+2014}{1+3+5+\dots+2013} - \frac{1+3+5+\dots+2013}{2+4+6+\dots+2014}. \] Find $m$. In other words, find the numerator of $A$ when $A$ is written as a fraction in simplest form. [i]Proposed by Evan Chen[/i]

Let $\mathbb N = {1, 2, 3, . . .}$. For real $x, y$, set $S(x, y) = \{s | s = [nx+y], n \in \mathbb N\}$. Prove that if $r > 1$ is a rational number, there exist real numbers $u$ and $v$ such that \[S(r, 0) \cap S(u, v) = \emptyset, S(r, 0) \cup S(u, v) = \mathbb N.\]