Team selection tests for JBMO 2018

We shall prove that for all $n>2$, Omar always has the strategy to win.

First, Ali has to merge some two bags of 1 candy into one bag of 2 candies. We prove that Omar can turn the state of all bags into: one bag contains odd number of candies and the others just have one candy each.

Indeed, in the second turn, Omar merges the bag of 2 candies with some bag of 1 candy to get one bag of 3 candies. Suppose that after a turn of Omar, there is a bag of $2k+1$ candies (with $k\in {\mathbb{Z}}^{+}$) and the other bags just have one candy each. At the next turn of Ali, there are two cases:

- If Ali merges two bags of 1 candy to one bag of 2 candies, then on the next turn, Omar merges that bag with the bag of $2k+1$ candies to get a bag of $2k+3$ candies.

- If Ali merges the bag of $2k+1$ candies with some bag of 1 candy to get another bag of $2k+2$, then on the next turn, Omar merges that bag with some bag of 1 candy to get a bag of $2k+3$ candies.

Hence, Omar always can control the state of bags like that. Note that after two turns of players, the number of bags reduces by 2. Finally, if after a turn of Omar, there is only one bag of $2k+1$ candies then Ali will be the loser. Otherwise, there is a bag of $2k+1$ candies and 3 bags of 1 candy. There are two cases:

- If Ali merges two bags of 1 candy then Omar merges the bag of $2k+1$ with the bag of 1 candy; then after that turn, there are a bag of 2 candies and a bag of $2k+2$ which implies that Ali loses.

- If Ali merges the bag of $2k+1$ candies with a bag of 1 candy then Omar merges two bags of 1 candy to make two bags of even number of candies as above.

Therefore, in all cases of $n>2$, Omar always can win the game.