Game of divisors: We have number N. On board we write down all divisiors of N. Players choose one of divisors. Lets say d. We erase d and all divisors of d. The player, who erase N, loose.
It was told me that player 2 have winnig strategy. Could you tell me how to show it? Any tips?
The first player can always win. The idea is kind of cute.
In the first turn player $A$ considers all of the integers $k>1$. If one of them can be deleted so that he has a winning strategy, he deletes it.
If none exist he just deletes $1$, and his opponent is now in a losing position.
If you look at the directed graph of the game what happens is that the vertex $I$ of the initial position is connected to a vertex $U$ (the one corresponding to deleting $1$). Such that the neighbours of $U$ are all neighbours of $I$. Every vertex $I$ with this property is a winning position.
Why? Because if $I$ where a losing position, so would be $U$, so $I$ is connected to a losing position, and hence $I$ would be a winning position, a contradiction.