60th Belarusian Mathematical Olympiad

Nick wins if he sets $n=2$. We consider all possibilities for $m$. Let $m=3$. If Mary removes $k$ stones then Nick removes $4-k$. In this case exactly $4$ stones are removed from the heap after each pair of moves (Mary - Nick). Since $360:4$, Nick wins.

Now we will solve the problem moving backward. We write all numbers from $1$ to $360$ and mark them with "+" or "-". If $k$ stones remain in the heap before the move of a player and he/she can win, then we write $+k$, otherwise we write $-k$. For $m=4,5,7,8$ we have the following table

| +1 | +2 | -3 | +4 | +5 | -6 | +7 | +8 | -9 | +10 | +11 | -12 | +13 | +14 | -15 | +16 | ... |

We see that the signs in the table are repeated with the period equaled $3$ (it is easy to prove by induction). Since $360:3$ it follows that $360$ is marked with the sign "-". Thus the number $360$ is the losing number of the stones for the beginning player.

For $m=6$ we have the table

| +1 | +2 | -3 | +4 | +5 | +6 | -7 | +8 | +9 | -10 | +11 | +12 | +13 | -14 | ... | | 1 | 2 | - | 1 | 2 | 6 | - | 1 | 2 | - | 1 | 2 | 6 | - | ... |

The second row of the table contains the possible winning moves (to win the player can remove so many stones that after his move the number of the stones remained in the heap would be marked with "+"). We see that the signs in the first row of the table are repeated with the period equaled $7$ (it is easy to prove by induction). Since $360$ is congruent to $3$ modulo $7$, it follows that $360$ and $3$ are marked with the same sign, i.e. "-". Thus the number $360$ is the losing number of the stones for the beginning player. To win Nick can remove so many stones as it is written in the second row of the table.

In a similar way, for $m=9$ we have the table

| +1 | +2 | -3 | +4 | +5 | -6 | +7 | +8 | +9 | -10 | | 1 | 2 | - | 1 | 2 | - | 1 | 2 | 9 | - |

| +11 | +12 | -13 | +14 | +15 | -16 | +17 | +18 | +19 | -20 | ... | | 1 | 2 | - | 1 | 2 | - | 1 | 2 | 9 | - | ... |

The second row of the table contains the possible winning moves (to win the player can remove so many stones that after his move the number of the stones remained in the heap would be marked with "+"). We see that the signs in the first row of the table are repeated with the period equaled $10$ (it is easy to prove by induction). Since $360$ is divided by $10$, it follows that $360$ and $10$ are marked with the same sign, i.e. "-". Thus the number $360$ is the losing number of the stones for the beginning player. To win Nick can remove so many stones as it is written in the second row of the table.