Game of chocolates

Question

Two players $A$ and $B$ play a game alternatively and $A$ starts the game.

Their are $2$ boxes of chocolates and we are Given the number of chocolates in both the boxes, let them be $c_1$ and $c_2$, the player takes either $c_1$ or $c_2$ number of chocolates and divide the remaining box of chocolates to two boxes (these two boxes need not have the same number of chocolates). The player who cannot make such a move loses.

Given the initial number of chocolates ($c_1$ and $c_2$) find the winner. Assume both the players play optimally.

Answer 1

Define a winning number to be one that can be divided into two nonzero parts $c_1$ and $c_2$ such that $(c_1, c_2)$ is a losing position. Clearly $(c_1, c_2)$ is a winning position when either $c_1$ or $c_2$ is a winning number; so $(c_1,c_2)$ is a losing position when $c_1$ and $c_2$ are both losing numbers; and we can say the following:

The sum of two losing numbers is a winning number.

Moreover, $1$ is a losing number. Hence $1+1=2$ is a winning number. Extending this pattern, we find that odd numbers are losing and even numbers are winning, and hence:

Given $(c_1, c_2)$, the first player can win if either $c_1$ or $c_2$ is even. In that case, he should leave an even number of chocolates uneaten, then divide them into two odd parts. Otherwise the first player loses.