Beat your Friend at Subtraction!

You should get an even integer in the beginning. Then, your strategy is to subtract $1$ from all the integers you are given. (You can do this because $1$ is a factor of every positive integer, except perhaps $1$ .) In this way, you will always give your friend odd integers, and your friend can only give you even integers, regardless of how your friend subtracts. This is because odd integers have only odd factors e.g. $15$ has proper factors $1, 3$ and $5$ . Since subtracting an odd from an odd yields an even, your friend is forced to give you even numbers.

Eventually, you will reach a point where you get the integer $2$ , and you give $2 - 1 = 1$ to your friend. You win the game. On the other hand, your friend can never win because $1$ is odd, and she can only produce even integers. (Conversely, if you are given an odd integer, then your friend can utilize the strategy to win, so you'll always lose.)

Thus, our solution is the number of even integers between $1$ and $2015$ . This is $\boxed{1007}$ .

The winning positions for the player who has the move are exactly the even integers.

Proof by complete induction:

Base step: $N = 1$

• By the rules of the game, this is losing.

Inductive step: Assume true for $N < K$ . For $N = K$ :

• If $K$ is even:

  • Player can subtract 1. By the induction hypothesis, $K - 1$ is losing, therefore $K$ is winning.

• If $K$ is odd:

  • Player can subtract only an odd number $L$ . The result will be a smaller even number $K - L$ . By the induction hypothesis, $K - L$ is winning for all available $L$ , therefore $K$ is losing.

There are $\boxed{1007}$ even integers in $[1, 2015]$ , therefore this is the number of winning positions.

Anyone care to try a variation? Everything is the same as the original question, except the loser is the first person to cause the number to drop below 6.