Find all four-digit numbers which are equal to the cube of the sum of their digits.

A positive integer is called [i]beautiful[/i] if it can be represented in the form $\dfrac{x^2+y^2}{x+y}$ for two distinct positive integers $x,y$. A positive integer that is not beautiful is [i]ugly[/i]. a) Prove that $2014$ is a product of a beautiful number and an ugly number. b) Prove that the product of two ugly numbers is also ugly.

If the remainder is $2013$ when a polynomial with coefficients from the set $\{0,1,2,3,4,5\}$ is divided by $x-6$, what is the least possible value of the coefficient of $x$ in this polynomial? $ \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ 1 $

The sweeties shop called "Olympiad" sells boxes of $6,9$ or $20$ chocolates. Groups of students from a school that is near the shop collect money to buy a chocolate for each student; to make this they buy a box and than give to everybody a chocolate. Like this students can create groups of $15=6+9$ students, $38=2*9+20$ students, etc. The seller has promised to the students that he can satisfy any group of students, and if he will need to open a new box of chocolate for any group (like groups of $4,7$ or $10$ students) than he will give all the chocolates for free to this group. Can there be constructed the biggest group that profits free chocolates, and if so, how many students are there in this group?

A bored student walks down a hall that contains a row of closed lockers, numbered 1 to 1024. He opens the locker numbered 1, and then alternates between skipping and opening each closed locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens?

Find all polynomials $ f\in\mathbb Z[x]$ such that for each $ a,b,x\in\mathbb N$ \[ a\plus{}b\plus{}c|f(a)\plus{}f(b)\plus{}f(c)\]

Find all integer solutions $(x,y)$ of the equation $y^2=x^3-p^2x,$ where $p$ is a prime such that $p\equiv 3 \mod 4.$

Let $\mathbb R$ be the real numbers. Set $S = \{1, -1\}$ and define a function $\operatorname{sign} : \mathbb R \to S$ by \[ \operatorname{sign} (x) = \begin{cases} 1 & \text{if } x \ge 0; \\ -1 & \text{if } x < 0. \end{cases} \] Fix an odd integer $n$. Determine whether one can find $n^2+n$ real numbers $a_{ij}, b_i \in S$ (here $1 \le i, j \le n$) with the following property: Suppose we take any choice of $x_1, x_2, \dots, x_n \in S$ and consider the values \begin{align*} y_i &= \operatorname{sign} \left( \sum_{j=1}^n a_{ij} x_j \right), \quad \forall 1 \le i \le n; \\ z &= \operatorname{sign} \left( \sum_{i=1}^n y_i b_i \right) \end{align*} Then $z=x_1 x_2 \dots x_n$.

Does there exist a sequence of natural numbers in which every natural number occurs exactly once, such that for each $k = 1, 2, 3, \dots$ the sum of the first $k$ terms of the sequence is divisible by $k$? [i]A. Shapovalov[/i]

Let $p$ be the largest prime less than $2013$ for which \[ N = 20 + p^{p^{p+1}-13} \] is also prime. Find the remainder when $N$ is divided by $10^4$. [i]Proposed by Evan Chen and Lewis Chen[/i]

The sum of two prime numbers is $85$. What is the product of these two prime numbers? $\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$

A $7 \times 1$ board is completely covered by $m \times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7 \times 1$ board in which all three colors are used at least once. For example, a $1 \times 1$ red tile followed by a $2 \times 1$ green tile, a $1 \times 1$ green tile, a $2 \times 1$ blue tile, and a $1 \times 1$ green tile is a valid tiling. Note that if the $2 \times 1$ blue tile is replaced by two $1 \times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$.

For a number $n$ in base $10$, let $f(n)$ be the sum of all numbers possible by removing some digits of $n$ (including none and all). For example, if $n = 1234$, $f(n) = 1234 + 123 + 124 + 134 + 234 + 12 + 13 + 14 + 23 + 24 + 34 + 1 + 2 + 3 + 4 = 1979$; this is formed by taking the sums of all numbers obtained when removing no digit from $n$ (1234), removing one digit from $n$ (123, 124, 134, 234), removing two digits from $n$ (12, 13, 14, 23, 24, 34), removing three digits from $n$ (1, 2, 3, 4), and removing all digits from $n$ (0). If $p$ is a 2011-digit integer, prove that $f(p)-p$ is divisible by $9$. Remark: If a number appears twice or more, it is counted as many times as it appears. For example, with the number $101$, $1$ appears three times (by removing the first digit, giving $01$ which is equal to $1$, removing the first two digits, or removing the last two digits), so it is counted three times.

Show that there exists a set $ A$ of positive integers with the following property: for any infinite set $ S$ of primes, there exist [i]two[/i] positive integers $ m$ in $ A$ and $ n$ not in $ A$, each of which is a product of $ k$ distinct elements of $ S$ for some $ k \geq 2$.

What is $3^{2002}$ in $\bmod 11$? $ \textbf{a)}\ 1 \qquad\textbf{b)}\ 3 \qquad\textbf{c)}\ 4 \qquad\textbf{d)}\ 5 \qquad\textbf{e)}\ \text{None of above} $

A [i]squiggle[/i] is composed of six equilateral triangles with side length $1$ as shown in the figure below. Determine all possible integers $n$ such that an equilateral triangle with side length $n$ can be fully covered with [i]squiggle[/i]s (rotations and reflections of [i]squiggle[/i]s are allowed, overlappings are not). [asy] import graph; size(100); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; draw((0,0)--(0.5,1),linewidth(2pt)); draw((0.5,1)--(1,0),linewidth(2pt)); draw((0,0)--(3,0),linewidth(2pt)); draw((1.5,1)--(2,0),linewidth(2pt)); draw((2,0)--(2.5,1),linewidth(2pt)); draw((0.5,1)--(2.5,1),linewidth(2pt)); draw((1,0)--(2,2),linewidth(2pt)); draw((2,2)--(3,0),linewidth(2pt)); dot((0,0),ds); dot((1,0),ds); dot((0.5,1),ds); dot((2,0),ds); dot((1.5,1),ds); dot((3,0),ds); dot((2.5,1),ds); dot((2,2),ds); clip((-4.28,-10.96)--(-4.28,6.28)--(16.2,6.28)--(16.2,-10.96)--cycle);[/asy]

How many real triples $(x,y,z)$ are there such that $x^4+y^4+z^4+1 = 4xyz$? $ \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ \text{Infinitely many} $

It is given that for some $n \in \mathbb{N}$ there exists a natural number $a$, such that $a^{n-1} \equiv 1 \pmod{n}$ and that for any prime divisor $p$ of $n-1$ we have $a^{\frac{n-1}{p}} \not \equiv 1 \pmod{n}$. Prove that $n$ is a prime.

Prove that if $m$ and $s$ are integers with $ms=2000^{2001}$, then the equation $mx^2-sy^2=3$ has no integer solutions.

Prove that for any n natural, the number \[ \sum \limits_{k=0}^{n} \binom{2n+1}{2k+1} 2^{3k} \] cannot be divided by $5$.

Find the sum of all integers $m$ with $1 \le m \le 300$ such that for any integer $n$ with $n \ge 2$, if $2013m$ divides $n^n-1$ then $2013m$ also divides $n-1$. [i]Proposed by Evan Chen[/i]

Let $a>1$ be an odd positive integer. Find the least positive integer $n$ such that $2^{2000}$ is a divisor of $a^n-1$. [i]Mircea Becheanu [/i]

Solve the equation \[ x^3+2y^3-4x-5y+z^2=2012, \] in the set of integers.

Prove that $\prod\limits_{k=0}^{n-1}{({{2}^{n}}-{{2}^{k}})}$ is divisible by $n!$ for all positive integers $n$.

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]