There are $n$ sheep and a wolf in sheep's clothing . Some of the sheep are friends (friendship is mutual). The goal of the wolf is to eat all the sheep. First, the wolf chooses some sheep to make friend's with. In each of the following days, the wolf eats one of its friends. Whenever the wolf eats a sheep $A$: (a) If a friend of $A$ is originally a friend of the wolf, it un-friends the wolf. (b) If a friend of $A$ is originally not a friend of the wolf, it becomes a friend of the wolf. Repeat the procedure until the wolf has no friend left. Find the largest integer $m$ in terms of $n$ satisfying the following: There exists an initial friendsheep structure such that the wolf has $m$ different ways of choosing initial sheep to become friends, so that the wolf has a way to eat all of the sheep.

The $n$ contestant of EGMO are named $C_1, C_2, \cdots C_n$. After the competition, they queue in front of the restaurant according to the following rules. [list] [*]The Jury chooses the initial order of the contestants in the queue. [*]Every minute, the Jury chooses an integer $i$ with $1 \leq i \leq n$. [list] [*]If contestant $C_i$ has at least $i$ other contestants in front of her, she pays one euro to the Jury and moves forward in the queue by exactly $i$ positions. [*]If contestant $C_i$ has fewer than $i$ other contestants in front of her, the restaurant opens and process ends. [/list] [/list] [list=a] [*]Prove that the process cannot continue indefinitely, regardless of the Jury’s choices. [*]Determine for every $n$ the maximum number of euros that the Jury can collect by cunningly choosing the initial order and the sequence of moves. [/list]

We have two lists of numbers: One initially containing 1,6,11,...,96, and the other initially containing 4,9,14,...,99. In every turn, we erase two numbers from one of the lists, and write $\frac{1}{3}$ of their sum (not necessarily an integer) in the other list. We continue this process until there are no possible moves. [list=a] [*] Prove that at the end of the process, there is exactly one number in each list. [*] Prove that those two numbers are [u]not[/u] equal. [/list]

Let $n\geq1$ be a positive integer. $n$ lamps are placed in a line. At minute 0, some lamps are on (maybe all of them). Every minute the state of the lamps changes: A lamp is on at minute $t+1$ if and only if at minute $t$, exactly one of its neighbors is on (the two lamps at the ends have one neighbor each, all other lamps have two neighbors). For which values of $n$ can we guarantee that all lamps will be off after some time?

There are 2010 students and 100 classrooms in the Olympiad High School. At the beginning, each of the students is in one of the classrooms. Each minute, as long as not everyone is in the same classroom, somebody walks from one classroom into a different classroom with at least as many students in it (prior to his move). This process will terminate in $M$ minutes. Determine the maximum value of $M$.

The numbers $1,2,\ldots,49,50$ are written on the blackboard. Ann performs the following operation: she chooses three arbitrary numbers $a,b,c$ from the board, replaces them by their sum $a+b+c$ and writes $(a+b)(b+c)(c+a)$ to her notebook. Ann performs such operations until only two numbers remain on the board (in total 24 operations). Then she calculates the sum of all $24$ numbers written in the notebook. Let $A$ and $B$ be the maximum and the minimum possible sums that Ann san obtain. Find the value of $\frac{A}{B}$. [i](I. Voronovich)[/i]

Simon plays a game on an $n\times n$ grid of cells. Initially, each cell is filled with an integer. Every minute, Simon picks a cell satisfying the following: [list] [*] The magnitude of the integer in the chosen cell is less than $n^{n^n}$ [*] The sum of all the integers in the neighboring cells (sharing one side with the chosen cell) is non-zero [/list] Simon then adds each integer in a neighboring cell to the chosen cell. Show that Simon will eventually not be able to make any valid moves.

Around a round table the people $P_1, P_2,..., P_{2013}$ are seated in a clockwise order. Each person starts with a certain amount of coins (possibly none); there are a total of $10000$ coins. Starting with $P_1$ and proceeding in clockwise order, each person does the following on their turn: [list][*]If they have an even number of coins, they give all of their coins to their neighbor to the left. [*]If they have an odd number of coins, they give their neighbor to the left an odd number of coins (at least $1$ and at most all of their coins) and keep the rest.[/list] Prove that, repeating this procedure, there will necessarily be a point where one person has all of the coins.

On the parliament of Sikinia, for any two deputies, there is third deputy, which knows exactly one of the two. Every deputy belongs to one of the two ruling parties. Every day, he president tells a certain group of deputies to change the party that they belong, and all the deputies which which know at least one of the deputies of the group has to change their party too. Prove that, the president can reach any configuration of deputies between two parties.(The president himself isn't a member of the parliament of Sikinia). [I]Proposed by S. Berlov[/i]