Given a fixed positive integer $a\geq 9$. Prove: There exist finitely many positive integers $n$, satisfying: (1)$\tau (n)=a$ (2)$n|\phi (n)+\sigma (n)$ Note: For positive integer $n$, $\tau (n)$ is the number of positive divisors of $n$, $\phi (n)$ is the number of positive integers $\leq n$ and relatively prime with $n$, $\sigma (n)$ is the sum of positive divisors of $n$.

Let $A$ be the set of $n$-digit integers whose digits are all from $\{ 1, 2, 3, 4, 5 \}$. $B$ is subset of $A$ such that it contains digit $5$, and there is no digit $3$ in front of digit $5$ (i.e. for $n = 2$, $35$ is not allowed, but $53$ is allowed). How many elements does set $B$ have?

Palindrome is a sequence of digits which doesn't change if we reverse the order of its digits. Prove that a sequence $(x_n)^{\infty}_{n=0}$ defined as $x_n=2013+317n$ contains infinitely many numbers with their decimal expansions being palindromes.

A series of numbers is called complete if it has non-zero natural terms and any nonzero integer has at least one among multiple series. Show that the arithmetic progression is a complete sequence if and only if it divides the first term relationship.

Prove that a finite simple planar graph has an orientation so that every vertex has out-degree at most 3.

Prove the following inequality: $x_1 + 2x_2 + 3x_3 + ... + nx_n \leq \frac{n(n-1)}{2} + x_1 + x_2 ^2 + x_3 ^3 + ... + x_n ^n$ where $\forall _{x_i} x_i > 0$

Let $(a_n)_n\geq 0$ and $a_{m+n}+a_{m-n}=\frac{1}{2}(a_{2m}+a_{2n})$ for every $m\geq n\geq0.$ If $a_1=1,$ then find the value of $a_{2007}.$

The Fibonacci sequence $\{F_{n}\}$ is defined by \[F_{1}=1, \; F_{2}=1, \; F_{n+2}=F_{n+1}+F_{n}.\] Show that $F_{mn-1}-F_{n-1}^{m}$ is divisible by $F_{n}^{2}$ for all $m \ge 1$ and $n>1$.

Consider a function $f: \mathbb Z \to \mathbb Z$ such that for every integer $n \ge 0$, there are at most $0.001n^2$ pairs of integers $(x,y)$ for which $f(x+y) \neq f(x)+f(y)$ and $\max\{ \lvert x \rvert, \lvert y \rvert \} \le n$. Is it possible that for some integer $n \ge 0$, there are more than $n$ integers $a$ such that $f(a) \neq a \cdot f(1)$ and $\lvert a \rvert \le n$? [i]Proposed by David Yang[/i]

Let $n$ be a fixed positive odd integer. Take $m+2$ [b]distinct[/b] points $P_0,P_1,\ldots ,P_{m+1}$ (where $m$ is a non-negative integer) on the coordinate plane in such a way that the following three conditions are satisfied: 1) $P_0=(0,1),P_{m+1}=(n+1,n)$, and for each integer $i,1\le i\le m$, both $x$- and $y$- coordinates of $P_i$ are integers lying in between $1$ and $n$ ($1$ and $n$ inclusive). 2) For each integer $i,0\le i\le m$, $P_iP_{i+1}$ is parallel to the $x$-axis if $i$ is even, and is parallel to the $y$-axis if $i$ is odd. 3) For each pair $i,j$ with $0\le i<j\le m$, line segments $P_iP_{i+1}$ and $P_jP_{j+1}$ share at most $1$ point. Determine the maximum possible value that $m$ can take.

Given is a function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $|f(x+y)-f(x)-f(y)|\leq 1$. Prove the existence of an additive function $g:\mathbb{R}\rightarrow \mathbb{R}$ (that is $g(x+y)=g(x)+g(y)$) such that $|f(x)-g(x)|\leq 1$ for any $x \in \mathbb{R}$

For $n\ge 1$ let $d_n$ be the $\gcd$ of the entries of $A^n-\mathcal{I}_2$ where \[ A=\begin{pmatrix} 3&2\\ 4&3\end{pmatrix}\quad \text{ and }\quad \mathcal{I}_2=\begin{pmatrix}1&0\\ 0&1\\\end{pmatrix}\] Show that $\lim_{n\to \infty}d_n=\infty$.

$a_1, a_2, ..., a_{95}$ are positive reals. Show that $\displaystyle \sum_{k=1}^{95}{a_k} \le 94+ \prod_{k=1}^{95}{\max{\{1,a_k\}}}$

Find all subsets $A$ of $\left\{ 1, 2, 3, 4, \ldots \right\}$, with $|A| \geq 2$, such that for all $x,y \in A, \, x \neq y$, we have that $\frac{x+y}{\gcd (x,y)}\in A$. [i]Dan Schwarz[/i]

(a) For each integer $k\ge 3$, find a positive integer $n$ that can be represented as the sum of exactly $k$ mutually distinct positive divisors of $n$. (b) Suppose that $n$ can be expressed as the sum of exactly $k$ mutually distinct positive divisors of $n$ for some $k\ge 3$. Let $p$ be the smallest prime divisor of $n$. Show that \[\frac1p+\frac1{p+1}+\cdots+\frac{1}{p+k-1}\ge1.\]