Problem (Combinatorics):
Players $A$ and $B$ play a game with $N \geq 2012$ coins and 2012 boxes arranged around a circle. Initially $A$ distributes the coins among the boxes so that there is at least 1 coin in each box. Then the two of them make moves in the order $B, A, B, A, \ldots$ by the following rules:

- On every move of his $B$ passes 1 coin from every box to an adjacent box.
- On every move of hers $A$ chooses several coins that were not involved in $B$ 's previous move and are in different boxes. She passes every chosen coin to an adjacent box.

Player $A$ 's goal is to ensure at least 1 coin in each box after every move of hers, regardless of how $B$ plays and how many moves are made. Find the least $N$ that enables her to succeed.

Solution:
We argue for a general $n \geq 7$ instead of 2012 and prove that the required minimum $N$ is $2 n-2$. For $n=2012$ this gives $N_{\min }=4022$.

a) If $N=2 n-2$ player $A$ can achieve her goal. Let her start the game with a regular distribution: $n-2$ boxes with 2 coins and 2 boxes with 1 coin. Call the boxes of the two kinds red and white respectively. We claim that on her first move $A$ can achieve a regular distribution again, regardless of $B$ 's first move $M$. She acts according as the following situation $S$ occurs after $M$ or not: The initial distribution contains a red box $R$ with 2 white neighbors, and $R$ receives no coins from them on move $M$.

Suppose that $S$ does not occur. Exactly one of the coins $c_{1}$ and $c_{2}$ in a given red box $X$ is involved in $M$, say $c_{1}$. If $M$ passes $c_{1}$ to the right neighbor of $X$, let $A$ pass $c_{2}$ to its left neighbor, and vice versa. By doing so with all red boxes $A$ performs a legal move $M^{\prime}$. Thus $M$ and $M^{\prime}$ combined move the 2 coins of every red box in opposite directions. Hence after $M$ and $M^{\prime}$ are complete each neighbor of a red box $X$ contains exactly 1 coin that was initially in $X$. So each box with a red neighbor is non-empty after $M^{\prime}$. If initially there is a box $X$ with 2 white neighbors ( $X$ is red and unique) then $X$ receives a coin from at least one of them on move $M$ since $S$ does not occur. Such a coin is not involved in $M^{\prime}$, so $X$ is also non-empty after $M^{\prime}$. Furthermore each box $Y$ has given away its initial content after $M$ and $M^{\prime}$. A red neighbor of $Y$ adds 1 coin to it; a white neighbor adds at most 1 coin because it is not involved in $M^{\prime}$. Hence each box contains 1 or 2 coins after $M^{\prime}$. Because $N=2 n-2$, such a distribution is regular.

Now let $S$ occur after move $M$. Then $A$ leaves untouched the exceptional red box $R$. With all remaining red boxes she proceeds like in the previous case, thus making a legal move $M^{\prime \prime}$. Box $R$ receives no coins from its neighbors on either move, so there is 1 coin in it after $M^{\prime \prime}$. Like above $M$ and $M^{\prime \prime}$ combined pass exactly 1 coin from every red box different from $R$ to each of its neighbors. Every box except $R$ has a red neighbor different from $R$, hence all boxes are non-empty after $M^{\prime \prime}$. Next, each box $Y$ except $R$ loses its initial content after $M$ and $M^{\prime \prime}$. A red neighbor of $Y$ adds at most 1 coin to it; a white neighbor also adds at most 1 coin as it does not participate in $M^{\prime \prime}$. Thus each box has 1 or 2 coins after $M^{\prime \prime}$, and the obtained distribution is regular.

Player $A$ can apply the described strategy indefinitely, so $N=2 n-2$ enables her to succeed.

b) For $N \leq 2 n-3$ player $B$ can achieve an empty box after some move of $A$. Let $\alpha$ be a set of $\ell$ consecutive boxes containing a total of $N(\alpha)$ coins. We call $\alpha$ an $\operatorname{arc}$ if $\ell \leq n-2$ and $N(\alpha) \leq 2 \ell-3$. Note that $\ell \geq 2$ by the last condition. Moreover if both extremes of $\alpha$ are non-empty boxes then $N(\alpha) \geq 2$, so that $N(\alpha) \leq 2 \ell-3$ implies $\ell \geq 3$. Observe also that if an extreme $X$ of $\alpha$ has more than 1 coin then ignoring $X$ yields a shorter arc. It follows that every arc contains an arc whose extremes have at most 1 coin each.

Given a clockwise labeling $1,2, \ldots, n$ of the boxes, suppose that boxes $1,2, \ldots, \ell$ form an arc $\alpha$, with $\ell \leq n-2$ and $N(\alpha) \leq 2 \ell-3$. Suppose also that all $n \geq 7$ boxes are non-empty. Then $B$ can move so that an arc $\alpha^{\prime}$ with $N\left(\alpha^{\prime}\right)<N(\alpha)$ will appear after any response of $A$.



One may assume exactly 1 coin in boxes 1 and $\ell$ by a previous remark. Let $B$ pass 1 coin in counterclockwise direction from box 1 and box $n$, and in clockwise direction from each remaining box. This leaves $N(\alpha)-2$ coins in the boxes of $\alpha$. In addition, due to $3 \leq \ell \leq n-2$, box $\ell$ has exactly 1 coin $c$, the one received from box $\ell-1$.

Let player $A$ 's next move $M$ pass $k \leq 2$ coins to boxes $1,2, \ldots, \ell$ from the remaining ones. Only boxes 1 and $\ell$ can receive such coins, at most 1 each. If $k<2$ then after move $M$ boxes $1,2, \ldots, \ell$ form an arc $\alpha^{\prime}$ with $N\left(\alpha^{\prime}\right)<N(\alpha)$. If $k=2$ then $M$ adds a coin to box $\ell$. Also $M$ does not move coin $c$ from $\ell$ because $c$ is involved in the previous move of $B$. In summary boxes $1,2, \ldots, \ell$ contain $N(\alpha)$ coins like before, so they form an arc. However there are 2 coins now in the extreme $\ell$ of the arc. Ignore $\ell$ to obtain a shorter arc $\alpha^{\prime}$ with $N\left(\alpha^{\prime}\right)<N(\alpha)$.

Consider any initial distribution without empty boxes. Since $N \leq 2 n-3$, there are at least 3 boxes in it with exactly 1 coin. It follows from $n \geq 7$ that some 2 of them are the extremes of an arc $\alpha$. Hence $B$ can make the move described above, which leads to an arc $\alpha^{\prime}$ with $N\left(\alpha^{\prime}\right)<N(\alpha)$ after $A^{\prime}$ 's response. If all boxes in the new distribution are non-empty he can repeat the same, and so on. Because $N(\alpha)$ cannot decrease indefinitely, an empty box will occur after some move of $A$.

Final answer: 4022