Proving two aspects of Nim For our game of Nim, we have $5$ piles. 1) Prove that if there is a pile with more stones than the Nim sum of all the other piles, then there is a move that makes the Nim sum equal to zero. 2) Prove that if the Nim sum is not zero, then one of the piles is bigger than the Nim sum of the all the other piles. Our thoughts so far: if a pile has more stones than the Nim sum of other piles, that means we have $x_1>x_2 \oplus x_3 \oplus x_4 \oplus x_5$. This hopefully means that $\exists$ a $1$ for $x_1$ in every bit where the binary sum on the right has a $1$ unless $x_1$ has a $1$ where the binary sum has a $0$; then, all subsequent bits of $x_1$ can be either $1$ or $0$. I want to show that $\exists t$ such that $x_1 \oplus ... \oplus(x_i-t)\oplus ...\oplus x_5 =0$. I feel like I am almost there for 1). Any help is greatly appreciated!

You’re making it too hard. Assuming that the piles are numbered so that $x_1$ is the one with more stones than the Nim sum of the other four, you have $x_1>x_2\oplus x_3\oplus x_4\oplus x_5$, so just remove

$$x_1-(x_2\oplus x_3\oplus x_4\oplus x_5)$$

stones from the first pile. This will leave $x_2\oplus x_3\oplus x_4\oplus x_5$ stones in the first pile, and $n\oplus n=0$ for all $n$.