Let $a$, $b$, $c$ be positive reals for which \begin{align*} (a+b)(a+c) &= bc + 2 \\ (b+c)(b+a) &= ca + 5 \\ (c+a)(c+b) &= ab + 9 \end{align*} If $abc = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$, compute $100m+n$. [i]Proposed by Evan Chen[/i]

Find all positive integers $n$ such that for all positive integers $m$, $1<m<n$, relatively prime to $n$, $m$ must be a prime number.

If $n$ is a natural number, prove that the number $(n+1)(n+2)\cdots(n+10)$ is not a perfect square.

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 $m + n.$

For every odd number $p>1$ we have: $\textbf{(A)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-2\qquad \textbf{(B)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p\\ \textbf{(C)}\ (p-1)^{\frac{1}{2}(p-1)} \; \text{is divisible by} \; p\qquad \textbf{(D)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p+1\\ \textbf{(E)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-1$

In a certain game, Auntie Hall has four boxes $B_1$, $B_2$, $B_3$, $B_4$, exactly one of which contains a valuable gemstone; the other three contain cups of yogurt. You are told the probability the gemstone lies in box $B_n$ is $\frac{n}{10}$ for $n=1,2,3,4$. Initially you may select any of the four boxes; Auntie Hall then opens one of the other three boxes at random (which may contain the gemstone) and reveals its contents. Afterwards, you may change your selection to any of the four boxes, and you win if and only if your final selection contains the gemstone. Let the probability of winning assuming optimal play be $\tfrac mn$, where $m$ and $n$ are relatively prime integers. Compute $100m+n$. [i]Proposed by Evan Chen[/i]

Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$. Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$. The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

For every natural number $ n $, define $ s(n) $ as the smallest natural number so that for every natural number $ a $ relatively prime to $n$, this equation holds: \[ a^{s(n)} \equiv 1 (mod n) \] Find all natural numbers $ n $ such that $ s(n) = 2010 $

Triangle $ABC$ is a right triangle with $AC=7,$ $BC=24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD=BD=15.$ Given that the area of triangle $CDM$ may be expressed as $\frac{m\sqrt{n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$

For every positive integer $ n$, let $ \phi(n)$ denotes the number of positive integers less than $ n$ that is relatively prime to $ n$ and $ \tau(n)$ denote the sum of all positive divisors of $ n$. Let $ n$ be a positive integer such that $ \phi(n)|n\minus{}1$ and that $ n$ is not a prime number. Prove that $ \tau(n)>2009$.

Prove that there exists an infinite number of relatively prime pairs $(m, n)$ of positive integers such that the equation \[x^3-nx+mn=0\] has three distint integer roots.

Prove that the set of integers of the form $2^{k}-3$ ($k=2,3,\cdots$) contains an infinite subset in which every two members are relatively prime.

Let $ n > 1$ be an integer, and $ n$ can divide $ 2^{\phi(n)} \plus{} 3^{\phi(n)} \plus{} \cdots \plus{} n^{\phi(n)},$ let $ p_{1},p_{2},\cdots,p_{k}$ be all distinct prime divisors of $ n$. Show that $ \frac {1}{p_{1}} \plus{} \frac {1}{p_{2}} \plus{} \cdots \plus{} \frac {1}{p_{k}} \plus{} \frac {1}{p_{1}p_{2}\cdots p_{k}}$ is an integer. ( where $ \phi(n)$ is defined as the number of positive integers $ \leq n$ that are relatively prime to $ n$.)

Six chairs sit in a row. Six people randomly seat themselves in the chairs. Each person randomly chooses either to set their feet on the floor, to cross their legs to the right, or to cross their legs to the left. There is only a problem if two people sitting next to each other have the person on the right crossing their legs to the left and the person on the left crossing their legs to the right. The probability that this will [b]not[/b] happen is given by $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

If $p$ and $q$ are natural numbers so that \[ \frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319}, \] prove that $p$ is divisible with $1979$.

For positive integers $a>b>1$, define \[x_n = \frac {a^n-1}{b^n-1}\] Find the least $d$ such that for any $a,b$, the sequence $x_n$ does not contain $d$ consecutive prime numbers. [i]V. Senderov[/i]