There are $ 2009$ pebbles in some points $ (x,y)$ with both coordinates integer. A operation consists in choosing a point $ (a,b)$ with four or more pebbles, removing four pebbles from $ (a,b)$ and putting one pebble in each of the points \[ (a,b\minus{}1),\ (a,b\plus{}1),\ (a\minus{}1,b),\ (a\plus{}1,b)\] Show that after a finite number of operations each point will necessarily have at most three pebbles. Prove that the final configuration doesn't depend on the order of the operations.

An underground burrow consists of an infinite sequence of rooms labeled by the integers $(\dots, -3, -2, -1, 0, 1, 2, 3,\dots)$. Initially, some of the rooms are occupied by one or more rabbits. Each rabbit wants to be alone. Thus, if there are two or more rabbits in the same room (say, room $m$), half of the rabbits (rounding down) will flee to room $m-1$, and half (also rounding down) to room $m+1$. Once per minute, this happens simultaneously in all rooms that have two or more rabbits. For example, if initially all rooms are empty except for $5$ rabbits in room $\#12$ and $2$ rabbits in room $\#13$, then after one minute, rooms $\text{\#11--\#14}$ will contain $2$, $2$, $2$, and $1$ rabbits, respectively, and all other rooms will be empty. Now suppose that initially there are $k+1$ rabbits in room $k$ for each $k=0, 1, 2, \ldots, 9, 10$, and all other rooms are empty. [list=a] [*]Show that eventually the rabbits will stop moving. [*] Determine which rooms will be occupied when this occurs. [/list]

We deal $n-1$ cards in some way to $n$ people sitting around a table. From then on, in one move a person with at least $2$ cards gives one card to each of his/her neighbors. Prove that eventually a state will be reached where everyone has at most one card.

There are 20 people at a party. Each person holds some number of coins. Every minute, each person who has at least 19 coins simultaneously gives one coin to every other person at the party. (So, it is possible that $A$ gives $B$ a coin and $B$ gives $A$ a coin at the same time.) Suppose that this process continues indefinitely. That is, for any positive integer $n$, there exists a person who will give away coins during the $n$th minute. What is the smallest number of coins that could be at the party? [i]Proposed by Ray Li[/i]

A finite number of coins are placed on an infinite row of squares. A sequence of moves is performed as follows: at each stage a square containing more than one coin is chosen. Two coins are taken from this square; one of them is placed on the square immediately to the left while the other is placed on the square immediately to the right of the chosen square. The sequence terminates if at some point there is at most one coin on each square. Given some initial configuration, show that any legal sequence of moves will terminate after the same number of steps and with the same final configuration.

There are $n$ boxes ${B_1},{B_2},\ldots,{B_n}$ from left to right, and there are $n$ balls in these boxes. If there is at least $1$ ball in ${B_1}$, we can move one to ${B_2}$. If there is at least $1$ ball in ${B_n}$, we can move one to ${B_{n - 1}}$. If there are at least $2$ balls in ${B_k}$, $2 \leq k \leq n - 1$ we can move one to ${B_{k - 1}}$, and one to ${B_{k + 1}}$. Prove that, for any arrangement of the $n$ balls, we can achieve that each box has one ball in it.

There are 20 people at a party. Each person holds some number of coins. Every minute, each person who has at least 19 coins simultaneously gives one coin to every other person at the party. (So, it is possible that $A$ gives $B$ a coin and $B$ gives $A$ a coin at the same time.) Suppose that this process continues indefinitely. That is, for any positive integer $n$, there exists a person who will give away coins during the $n$th minute. What is the smallest number of coins that could be at the party? [i]Proposed by Ray Li[/i]