2015
Problem - 2043
Chad has $100$ cookies that he wants to distribute among four friends. Two of them, Jeff and Qiao, are rivals; neither wants the other to receive more cookies than they do. The other two, Jim and Townley, don't care about how many cookies they receive. In how many ways can Chad distribute all $100$ cookies to his four friends so that everyone is satisfied? (Some of his four friends may receive zero cookies.)
This is equivalent to finding the number of non-negative integer solutions to the following equation: $$2x+y+z=100$$
It is clear that $x$ must satisfy $0 \le x\le 50$. For each $x$, there will be $(100-2x)$ cookies left to be distributed among the other two people. It is easy to show there are $(101-2x)$ ways because one of them can receive any amount from $0$ to $(100-2x)$ which uniquely determines the amount the other can get. Therefore the answer is $$\sum_{x=0}^{50}(101-2x)=51\times 101 - 2\times\frac{50\times 51}{2}=\boxed{2601} $$