2. $[\mathbf{1 5}]$ Consider the following two-player game. Player 1 starts with a number, $N$. He then subtracts a proper divisor of $N$ from $N$ and gives the result to player 2 (a proper divisor of $N$ is a positive divisor of $N$ that is not equal to 1 or $N$ ). Player 2 does the same thing with the number she gets from player 1 , and gives the result back to player 1. The two players continue until a player is given a prime number or 1 , at which point that player loses. For which values of $N$ does player 1 have a winning strategy?
Answer: All even numbers except for odd powers of 2 . First we show that if you are stuck with an odd number, then you are guaranteed to lose. Suppose you have an odd number $a b$, where $a$ and $b$ are odd numbers, and you choose to subtract $a$. You pass your opponent the number $a(b-1)$. This cannot be a power of 2 (otherwise $a$ is a power of 2 and hence $a=1$, which is not allowed), so your opponent can find an odd proper divisor of $a(b-1)$, and you will have a smaller odd number. Eventually you will get to an odd prime and lose.
Now consider even numbers that aren't powers of 2 . As with before, you can find an odd proper divisor of $N$ and pass your opponent an odd number, so you are guaranteed to win.
Finally consider powers of 2 . If you have the number $N=2^{k}$, it would be unwise to choose a proper divisor other than $2^{k-1}$; otherwise you would give your opponent an even number that isn't a power of 2. Therefore if $k$ is odd, you will end up with 2 and lose. If $k$ is even, though, your opponent will end up with 2 and you will win.
Therefore player 1 has a winning strategy for all even numbers except for odd powers of 2 .