Prove that for all odd $ n > 1$, we have $ 8n \plus{} 4|C^{4n}_{2n}$.

Find all functions $f$ from the set of real numbers to itself satisfying \[ f(x(1+y)) = f(x)(1 + f(y)) \] for all real numbers $x, y$.

The infinite sequence $P_0(x),P_1(x),P_2(x),\ldots,P_n(x),\ldots$ is defined as \[P_0(x)=x,\quad P_n(x) = P_{n-1}(x-1)\cdot P_{n-1}(x+1),\quad n\ge 1.\] Find the largest $k$ such that $P_{2014}(x)$ is divisible by $x^k$.

Lucía multiplies some positive one-digit numbers (not necessarily distinct) and obtains a number $n$ greater than 10. Then, she multiplies all the digits of $n$ and obtains an odd number. Find all possible values of the units digit of $n$. $\textit{Proposed by Pablo Serrano, Ecuador}$

Let $\mathbb N$ be the set of positive integers. Find all functions $f\colon\mathbb N\to \mathbb N$ such that for every $m,n\in \mathbb N$, \[ f(m)+f(n)\mid m+n. \]

Let $m$ be a positive integer, and consider a $m\times m$ checkerboard consisting of unit squares. At the centre of some of these unit squares there is an ant. At time $0$, each ant starts moving with speed $1$ parallel to some edge of the checkerboard. When two ants moving in the opposite directions meet, they both turn $90^{\circ}$ clockwise and continue moving with speed $1$. When more than $2$ ants meet, or when two ants moving in perpendicular directions meet, the ants continue moving in the same direction as before they met. When an ant reaches one of the edges of the checkerboard, it falls off and will not re-appear. Considering all possible starting positions, determine the latest possible moment at which the last ant falls off the checkerboard, or prove that such a moment does not necessarily exist. [i]Proposed by Toomas Krips, Estonia[/i]

Find all pairs of integers $a,b$ for which there exists a polynomial $P(x) \in \mathbb{Z}[X]$ such that product $(x^2+ax+b)\cdot P(x)$ is a polynomial of a form \[ x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0 \] where each of $c_0,c_1,\ldots,c_{n-1}$ is equal to $1$ or $-1$.

Find all functions $f: Q^{+}\rightarrow R$ such that $f(x)+f(y)+2xyf(xy)=\frac{f(xy)}{f(x+y)}$ for all $x,y\in Q^{+}$

Let $n\geq3$ be an integer. We say that an arrangement of the numbers $1$, $2$, $\dots$, $n^2$ in a $n \times n$ table is [i]row-valid[/i] if the numbers in each row can be permuted to form an arithmetic progression, and [i]column-valid[/i] if the numbers in each column can be permuted to form an arithmetic progression. For what values of $n$ is it possible to transform any row-valid arrangement into a column-valid arrangement by permuting the numbers in each row?

A directed graph has each vertex with outdegree 2. Prove that it is possible to split the vertices into 3 sets so that for each vertex $v$, $v$ is not simultaneously in the same set with both of the vertices that it points to. [i]David Yang.[/i] [hide="Stronger Version"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=492100]here[/url].[/hide]

Define $a_1=1$ and $a_{n+1}=\frac{a_n}{n}+\frac{n}{a_n}$ for $n\in\mathbb{N}$. Find the greatest integer not exceeding $a_{2000}$ and prove your claim.

To [i]clip[/i] a convex $n$-gon means to choose a pair of consecutive sides $AB, BC$ and to replace them by the three segments $AM, MN$, and $NC$, where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$. In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$-gon. A regular hexagon ${\cal P}_6$ of area 1 is clipped to obtain a heptagon ${\cal P}_7$. Then ${\cal P}_7$ is clipped (in one of the seven possible ways) to obtain an octagon ${\cal P}_8$, and so on. Prove that no matter how the clippings are done, the area of ${\cal P}_n$ is greater than $\frac 13$, for all $n \geq 6$.

Let $({{x}_{n}}),({{y}_{n}})$ be two positive sequences defined by ${{x}_{1}}=1,{{y}_{1}}=\sqrt{3}$ and \[ \begin{cases} {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\ x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \] for all $n=1,2,3,\ldots$. Prove that they are converges and find their limits.

For any integer $d > 0,$ let $f(d)$ be the smallest possible integer that has exactly $d$ positive divisors (so for example we have $f(1)=1, f(5)=16,$ and $f(6)=12$). Prove that for every integer $k \geq 0$ the number $f\left(2^k\right)$ divides $f\left(2^{k+1}\right).$ [i]Proposed by Suhaimi Ramly, Malaysia[/i]