Given a paper on which the numbers $1,2,3\dots ,14,15$ are written. Andy and Bobby are bored and perform the following operations, Andy chooses any two numbers (say $x$ and $y$) on the paper, erases them, and writes the sum of the numbers on the initial paper. Meanwhile, Bobby writes the value of $xy(x+y)$ in his book. They were so bored that they both performed the operation until only $1$ number remained. Then Bobby adds up all the numbers he wrote in his book, let’s call $k$ as the sum. $a$. Prove that $k$ is constant which means it does not matter how they perform the operation, $b$. Find the value of $k$.

We are given $2n$ natural numbers \[1, 1, 2, 2, 3, 3, \ldots, n - 1, n - 1, n, n.\] Find all $n$ for which these numbers can be arranged in a row such that for each $k \leq n$, there are exactly $k$ numbers between the two numbers $k$.

Initially, on a board there a positive integer. If board contains the number $x,$ then we may additionally write the numbers $2x+1$ and $\frac{x}{x+2}.$ At some point 2008 is written on the board. Prove, that this number was there from the beginning.

We are given an $n \times n$ board, where $n$ is an odd number. In each cell of the board either $+1$ or $-1$ is written. Let $a_k$ and $b_k$ denote them products of numbers in the $k^{th}$ row and in the $k^{th}$ column respectively. Prove that the sum $a_1 +a_2 +\cdots+a_n +b_1 +b_2 +\cdots+b_n$ cannot be equal to zero.

Around a circular table an even number of persons have a discussion. After a break they sit again around the circular table in a different order. Prove that there are at least two people such that the number of participants sitting between them before and after a break is the same.

Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers $x$ and $y$ such that $x>y$ and $x$ is to the left of $y$, and replaces the pair $(x,y)$ by either $(y+1,x)$ or $(x-1,x)$. Prove that she can perform only finitely many such iterations. [i]Proposed by Warut Suksompong, Thailand[/i]

A little boy wrote the numbers $1,2,\cdots,2011$ on a blackboard. He picks any two numbers $x,y$, erases them with a sponge and writes the number $|x-y|$. This process continues until only one number is left. Prove that the number left is even.

For two real numbers $ a$, $ b$, with $ ab\neq 1$, define the $ \ast$ operation by \[ a\ast b=\frac{a+b-2ab}{1-ab}.\] Start with a list of $ n\geq 2$ real numbers whose entries $ x$ all satisfy $ 0<x<1$. Select any two numbers $ a$ and $ b$ in the list; remove them and put the number $ a\ast b$ at the end of the list, thereby reducing its length by one. Repeat this procedure until a single number remains. $ a.$ Prove that this single number is the same regardless of the choice of pair at each stage. $ b.$ Suppose that the condition on the numbers $ x$ is weakened to $ 0<x\leq 1$. What happens if the list contains exactly one $ 1$?

A hexagonal table is given, as the one on the drawing, which has $~$ $2012$ $~$ columns. There are $~$ $2012$ $~$ hexagons in each of the odd columns, and there are $~$ $2013$ $~$ hexagons in each of the even columns. The number $~$ $i$ $~$ is written in each hexagon from the $~$ $i$-th column. Changing the numbers in the table is allowed in the following way: We arbitrarily select three adjacent hexagons, we rotate the numbers, and if the rotation is clockwise then the three numbers decrease by one, and if we rotate them counterclockwise the three numbers increase by one (see the drawing below). What's the maximum number of zeros that can be obtained in the table by using the above-defined steps.

Given a positive integer $k$ and other two integers $b > w > 1.$ There are two strings of pearls, a string of $b$ black pearls and a string of $w$ white pearls. The length of a string is the number of pearls on it. One cuts these strings in some steps by the following rules. In each step: [b](i)[/b] The strings are ordered by their lengths in a non-increasing order. If there are some strings of equal lengths, then the white ones precede the black ones. Then $k$ first ones (if they consist of more than one pearl) are chosen; if there are less than $k$ strings longer than 1, then one chooses all of them. [b](ii)[/b] Next, one cuts each chosen string into two parts differing in length by at most one. (For instance, if there are strings of $5, 4, 4, 2$ black pearls, strings of $8, 4, 3$ white pearls and $k = 4,$ then the strings of 8 white, 5 black, 4 white and 4 black pearls are cut into the parts $(4,4), (3,2), (2,2)$ and $(2,2)$ respectively.) The process stops immediately after the step when a first isolated white pearl appears. Prove that at this stage, there will still exist a string of at least two black pearls. [i]Proposed by Bill Sands, Thao Do, Canada[/i]

The cirumcentre of the cyclic quadrilateral $ABCD$ is $O$. The second intersection point of the circles $ABO$ and $CDO$, other than $O$, is $P$, which lies in the interior of the triangle $DAO$. Choose a point $Q$ on the extension of $OP$ beyond $P$, and a point $R$ on the extension of $OP$ beyond $O$. Prove that $\angle QAP=\angle OBR$ if and only if $\angle PDQ=\angle RCO$.

Two people play a game as follows: At the beginning both of them have one point and in every move, one of them can double it's points, or when the other have more point than him, subtract to him his points. Can the two competitors have 2009 and 2002 points respectively? What about 2009 and 2003? Generally which couples of points can they have?

On the table lay a pencil, sharpened at one end. The student can rotate the pencil around one of its ends at $45^{\circ}$ clockwise or counterclockwise. Can the student, after a few turns of the pencil, go back to the starting position so that the sharpened end and the not sharpened are reversed?

There are three boxes of stones. Sisyphus moves stones one by one between the boxes. Whenever he moves a stone, Zeus gives him the number of coins that is equal to the difference between the number of stones in the box the stone was put in, and that in the box the stone was taken from (the moved stone does not count). If this difference is negative, then Sisyphus returns the corresponding amount to Zeus (if Sisyphus cannot pay, generous Zeus allows him to make the move and pay later). After some time all the stones lie in their initial boxes. What is the greatest possible earning of Sisyphus at that moment? [i]I. Izmest’ev[/i]

Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers $x$ and $y$ such that $x>y$ and $x$ is to the left of $y$, and replaces the pair $(x,y)$ by either $(y+1,x)$ or $(x-1,x)$. Prove that she can perform only finitely many such iterations. [i]Proposed by Warut Suksompong, Thailand[/i]

If $a,b,c,d \in \mathbb{R}_{+}$ and $a+b +c +d =1$, show that \[ ab +bc +cd \leq \dfrac{1}{4}. \]

Consider a $2^k$-tuple of numbers $(a_1,a_2,\dots,a_{2^k})$ all equal to $1$ or $-1$. In one step, we transform it to $(a_1a_2,a_2a_3,\dots,a_{2^k}a_1)$. Prove that eventually, we will obtain a $2^k$-tuple consisting only of $1$'s.

We have $2^m$ sheets of paper, with the number $1$ written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a + b$ on both sheets. Prove that after $m2^{m -1}$ steps, the sum of the numbers on all the sheets is at least $4^m$ . [i]Proposed by Abbas Mehrabian, Iran[/i]

Let $a,b$ be two positive integers, such that $ab\neq 1$. Find all the integer values that $f(a,b)$ can take, where \[ f(a,b) = \frac { a^2+ab+b^2} { ab- 1} . \]

Five identical empty buckets of $2$-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighbouring buckets, empties them to the river and puts them back. Then the next round begins. The Stepmother goal's is to make one of these buckets overflow. Cinderella's goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow? [i]Proposed by Gerhard Woeginger, Netherlands[/i]

Let $ k$ be an integer, $ k \geq 2$, and let $ p_{1},\ p_{2},\ \ldots,\ p_{k}$ be positive reals with $ p_{1} \plus{} p_{2} \plus{} \ldots \plus{} p_{k} \equal{} 1$. Suppose we have a collection $ \left(A_{1,1},\ A_{1,2},\ \ldots,\ A_{1,k}\right)$, $ \left(A_{2,1},\ A_{2,2},\ \ldots,\ A_{2,k}\right)$, $ \ldots$, $ \left(A_{m,1},\ A_{1,2},\ \ldots,\ A_{m,k}\right)$ of $ k$-tuples of finite sets satisfying the following two properties: (i) for every $ i$ and every $ j \neq j^{\prime}$, $ A_{i,j}\cap A_{i,j^{\prime}} \equal{} \emptyset$, and (ii) for every $ i\neq i^{\prime}$ there exist $ j\neq j^{\prime}$ for which $ A_{i,j} \cap A_{i^{\prime},j^{\prime}}\neq\emptyset$. Prove that \[ \sum_{b \equal{} 1}^{m}{\prod_{a \equal{} 1}^{k}{p_{a}^{|A_{b,a}|}}} \leq 1. \]

Let $ n \geq 2$ be a positive integer and $ \lambda$ a positive real number. Initially there are $ n$ fleas on a horizontal line, not all at the same point. We define a move as choosing two fleas at some points $ A$ and $ B$, with $ A$ to the left of $ B$, and letting the flea from $ A$ jump over the flea from $ B$ to the point $ C$ so that $ \frac {BC}{AB} \equal{} \lambda$. Determine all values of $ \lambda$ such that, for any point $ M$ on the line and for any initial position of the $ n$ fleas, there exists a sequence of moves that will take them all to the position right of $ M$.

Let $n > 3$ be a positive integer. Suppose that $n$ children are arranged in a circle, and $n$ coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their immediate neighbors on the right and left. Determine all initial distributions of the coins from which it is possible that, after a finite number of steps, each child has exactly one coin.

Let $n > 3$ be a positive integer. Suppose that $n$ children are arranged in a circle, and $n$ coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their immediate neighbors on the right and left. Determine all initial distributions of the coins from which it is possible that, after a finite number of steps, each child has exactly one coin.

A crazy physicist discovered a new kind of particle wich he called an imon, after some of them mysteriously appeared in his lab. Some pairs of imons in the lab can be entangled, and each imon can participate in many entanglement relations. The physicist has found a way to perform the following two kinds of operations with these particles, one operation at a time. (i) If some imon is entangled with an odd number of other imons in the lab, then the physicist can destroy it. (ii) At any moment, he may double the whole family of imons in the lab by creating a copy $I'$ of each imon $I$. During this procedure, the two copies $I'$ and $J'$ become entangled if and only if the original imons $I$ and $J$ are entangled, and each copy $I'$ becomes entangled with its original imon $I$; no other entanglements occur or disappear at this moment. Prove that the physicist may apply a sequence of such operations resulting in a family of imons, no two of which are entangled.