Game Of Partitions

The answer ends up being a bit simpler than you might think. The optimal solution for both players is to simply split each number $n$ into $1$ and $n-1$ , starting with $N$ and ending with $2$ being split into two $1$ s. After $N-1$ turns the game is concludes, no matter how the game is played.

If $N$ is even, then the number of rounds will be odd, meaning that Carla will get an extra turn. Each player gets $1$ point per turn and then Carla will get an extra 2 points for the last split, meaning that the score will end $\frac{N}{2}+2$ to $\frac{N}{2}$ in favor of Carla.

If $N$ is odd, then the number of rounds will be even, but since Alice makes the last move, she will get an extra point for making the last split. This gives us a final score of $\frac{N-1}{2}$ to $\frac{N+1}{2}$ in favor of Alice.

There are $50$ even integers less than or equal to 100, therefore Carla has $\boxed{50}$ numbers she can choose from to win the game.

It is easy to see from experimentation that Carla wins when N is even and Alice wins when N is odd. Proving this, however, is an interesting problem.

Start with the two simplest cases, when N = 2 and N = 3. Clearly, Carla will win in the first case, and Alice in the second. From here, we can induct upward. Suppose that for some M, we know that all N < M yields a win for Carla for even N, and a win for Alice for odd N.

Now consider the case where we begin with M. If M is even, Carla's strategy is to split it into 1 and M-1. M-1 is an odd number less than M, meaning that the first person who splits it will get less 1's. Since Alice has to start splitting M-1 first, and there is nothing else to split, she will get less 1's than Carla from the M-1, no matter what she does. This, combined with the fact that Carla got a 1 earlier, guarantees Carla will win for even M .

It gets a little trickier when M is odd. Carla's first move will split M into two numbers, denoted by A and B, of opposite parity. WLOG, suppose that A > B. Alice then splits A into B and C, where C = A-B. Because A and B are of opposite parity, C must be odd. Carla is then left with B, C, B (consider everything from here to be an action on one of these "trees").

From here, Alice's strategy is as follows: every time Carla splits a B-tree, Alice will do the exact same thing on the other B-tree. Thus, Carla must be the first person to split the C-tree. Since C < M and Carla split C first, Alice has the winning strategy on C, so she will employ the appropriate move on C every time Carla splits C. Ultimately, Alice will gain more from the C-tree, and they will both gain the same amounts from the B-trees because they do the exact same moves. Thus, Alice wins for odd M .

There are 50 even numbers between 2 and 100, inclusive, so the answer is $\boxed{50}$