Answer:
To determine the winning strategy for the first player in this game, we can start by looking at small values of n and see if we can identify a pattern.
For n = 1, the first player must remove the only token, so they win.
For n = 2, the first player can remove one token and win, or they can remove two tokens and lose.
For n = 3, the first player can remove one token, leaving two tokens for the second player, who will then be forced to lose.
For n = 4, the first player can remove two tokens, leaving two tokens for the second player, who will then be forced to lose.
Based on these observations, it seems that if n is a multiple of 3, the first player has a winning strategy. To see why this is the case, we can use mathematical induction.
Assume that if n is a multiple of 3, the first player has a winning strategy. We want to show that if n+1 is a multiple of 3, the first player also has a winning strategy.
If n+1 is a multiple of 3, we can write n+1 = 3k+1 for some integer k.
If the first player removes one token, there will be n tokens left, and the second player will be in a losing position by our assumption.
If the first player removes two tokens, there will be n-1 tokens left. If the second player removes one token, there will be n-2 tokens left, and the first player will be in a winning position by our assumption. If the second player removes two tokens, there will be n-3 tokens left, and the first player can remove one token, leaving n-4 tokens for the second player. At this point, the second player will be in a losing position by our assumption.
Therefore, we have shown that if n is a multiple of 3, the first player has a winning strategy, and this strategy can be described as follows:
If n is already a multiple of 3, remove one token.
If n is not a multiple of 3, remove two tokens if possible, otherwise remove one token.
We can also observe that if n is not a multiple of 3, the second player has a winning strategy.
Explanation: