I have a question regarding using strong induction to prove a concept relevant to the Game of Nim.
Here is the question: The game of Nim is a two player game where each player takes turns removing either 1,2 or 3 coins from a pile of coins. The player who removes the last coin loses. Whenever the number of coins is $m \in \mathbb{N}$ such that $m \equiv 1 \ (mod \ 4)$ (That is, $m=4n + 1$ for a non negative integer $n$) the second player has a winning strategy no matter what the first player does. Use strong induction on $n$ to prove this.
So far I've got that if the number of coins is not a multiple of $4$ the second player has a winning strategy and that the number of groups is $4$ is what we are investigating in the induction, not the number of coins.
Thanks in advance.