9.6. Two players play the following game. They have a chocolate bar, divided by grooves parallel to the sides of the bar into pieces. The grooves divide the bar into M vertical and N horizontal strips. The first player, on their turn, breaks the bar along one of the grooves into two rectangular parts and gives them to the second player. The second player chooses one of the parts, eats it, and breaks the other part along a groove, giving the resulting two parts to the first player. The first player eats one of the received parts, breaks the other, and gives it to the second player, and the process repeats. The player who cannot make a move loses. For which M and N can the first player play in such a way as to win regardless of the second player's actions?
Answer: The maximum power of 2 that divides $M$ does not coincide with that for $\mathrm{N}$.

Criteria

+ correct solution.

+/- correct answer and the beginning of the proof, including the case of a tile $2 n \times(2 m+1)$, but not limited to it.

+/2 cases of tiles $2 n \times(2 m+1)$ and $4 n \times(4 m+2)$ are considered.

+/3 it is proven that a tile of size $2 n \times(2 m+1)$ or $n \times 2 n-$ is suitable.

-/+ the division of all possible tile sizes into winning and losing ones has begun; or the same as +/3, with minor flaws; or the case of a tile $n \times(2 n+1)$ is considered.

-/ . a finite number of non-trivial cases leading to a win, greater than 3, is considered; or the case of a tile $1 \times 2 n$ is considered.

Solution. Let's call a tile losing if the maximum power of 2 that divides its width coincides with the maximum power of 2 that divides its height. Note that a losing tile cannot be broken into two losing ones: indeed, if the tile consists of two losing parts, then their common side has a length of $2^{s}(2 k+1)$, and the remaining sides are $2^{s}(2 m+1)$ and $2^{s}(2 n+1)$, so the entire tile has dimensions $2^{s}(2 k+1)$ and $2^{s+1}(m+n+1)$ and is not losing.

If a player receives a losing tile (at the beginning of the game) or two losing tiles (from the opponent), then, whichever of them he eats, when breaking the remaining losing tile, at least one of the pieces he gives to the opponent will be non-losing (see the note above). If, however, a player receives at least one non-losing tile (at the beginning of the game or from the opponent), that is, a tile of size $2^{s}(2 k+1)$ by $2^{t}(2 m+1)$ with $s>t$, then he can always break it into two losing ones, separating a piece of size $2^{t}$ by $2^{t}(2 m+1)$. Let's call this rule of breaking the tile a winning one.

Thus, if the initial tile is not losing, the first player can follow the winning rule every move, so the second player will always receive only losing tiles from him, and therefore will be forced to give the first player at least one winning tile. In less than $\mathrm{M}+\mathrm{N}$ moves, the size of the largest of the given tiles will drop to 1 by 1, and a pair of such tiles will go to the second player, because both are losing. At this moment, the second player will lose.

Similarly, if the initial tile was losing, then the first player will give the second at least one non-losing tile every move, and the second player can apply the winning rule every move, as a result of which in less than $\mathrm{M}+\mathrm{N}$ moves a pair of tiles of size 1 by 1 will go to the first player and he will lose.

Second method. Note that each tile has one of two properties. Either the tile is winning, or it is losing. A winning tile is one that can be broken in such a way that regardless of the opponent's actions, it will lead to the victory of the one breaking the tile. A losing tile is one where any break leads to a loss. A winning tile can be broken into two losing ones. Accordingly, when breaking a losing tile, at least one of the tiles will be winning.

Let's start looking for winning tiles. Suppose a tile has the form $1 \times m$. Consider winning positions. The tile $1 \times 1$ is losing, $1 \times 2$ is winning, $1 \times 3$ is losing, $1 \times 4$ can be broken into $1 \times 1$ and $1 \times 3$ and the breaker wins. If the tile $1 \times m$ is losing, then the tile $1 \times(m+1)$ is winning, since it can be broken into pieces $1 \times m$ and $1 \times 1$. Let $1 \times m$ be a winning tile. Up to $m$, winning and losing tiles alternated, so all winning tiles have an even number of squares, and $m$ is even, while $m+1$ is odd. But an odd number cannot be split into the sum of two odd numbers, so $m+1$ cannot be broken into two losing tiles, meaning it is losing.

Note that for the case $(2 l+1) \times 2 k$ the same strategy works as for the case $1 \times 2 k$. Break off a piece from the side $2 k$. The second player will receive tiles $(2 l+1) \times 1$ and $(2 l+1) \times (2 k-1)$. The first is losing, so he breaks the second, but after any break at least one tile will have an even side. The first player chooses this one and repeats the actions. Thus, on the one hand, he can always make a move described by the strategy, and on the other hand, he will never receive two tiles $1 \times 1$ since they both have odd sides. Incidentally, we have proven that all tiles of the form $(2 k+1) \times(2 l+1)$ are losing.

Now consider the case when both sides are even. Suppose a tile of size $n \times m$ is winning (losing). Then a tile $2 n \times 2 m$ is also winning (losing).

We will prove this statement by induction on $(m+n)$. Base cases $m+n=2$ and $m+n=3$: tiles $(1,1)$ and $(2,2)$ are losing (case $m+n=2$), while tiles $(1,2)$ and $(2,4)$ are winning.

Inductive hypothesis: suppose the statement is true for all $m+n \leq k$. Prove for $m+n=k+1$.

First consider the case of a winning tile. Without loss of generality, we can assume that the first winning move for the tile $(n \times m)$ is to cut off a piece of size $x$ from the side $n$. In this case, the tile is divided into parts $x \times m$ and $(n-x) \times m$ and both tiles are losing. In the new tile, cut off a piece of size $2 x$ from $2 n$, the tile will be divided into parts of size $2 x \times 2 m$ and $2(n-x) \times 2 m$ which are proportional to $x \times m$ and $(n-x) \times m$ respectively, and therefore by the inductive hypothesis have the same outcomes. Tiles $x \times m$ and $(n-x) \times m$ are losing, so tiles $2 x \times 2 m$ and $2(n-x) \times 2 m$ are also losing.

Now consider the case of a losing tile. Then for any break of the tile $m \times n$, one of the tiles in the pair will be winning. Suppose the first player divides the new tile into parts $x \times 2 m$ and $(2 n-x) \times 2 m$. Then either $x$ is even and $x=2 x_{0}$, or $x$ is odd. In the first case, the resulting tiles have both sides even. By the inductive hypothesis, their outcomes are the same as in the case of $x_{0} \times m$ and $\left(n-x_{0}\right) \times m$. Tiles $x_{0} \times m$ and $\left(n-x_{0}\right) \times m$ are obtained by cutting off a part $x_{0}$ from the tile $n \times m$. The tile $n \times m$ is losing, so one of the tiles $x_{0} \times m$ and $\left(n-x_{0}\right) \times m$ is winning. In the second case, two tiles of the form $(2 k+1) \times 2 m$ are obtained, both of which are winning. The inductive step is proven.

Suppose the first player receives a tile of size $2^{k_{1}} n \times 2^{k_{2}} m$, where $k_{1} \leq k_{2}$ and the numbers $m, n$ are odd. The outcome of the game with such a tile is the same as the outcome of the game with a tile $n \times 2^{k_{1}-k_{2}} m$. The outcome for a tile with one odd side is determined by the parity of the second side. Accordingly, if $k_{1}=k_{2}$, the first player will lose. In other cases, the first player wins.