Game of Nim Sum Challenge Problem

Question

Suppose in the game of Nim there are 72 chips in the first pile, 60 chips in the second pile, and 100 chips in the third pile and it is your turn to play. How would you play?

Following the below formula for base two, how can I go about solving?

Answer 1

$72\oplus 60 \oplus 100 = 16$ as a decent binary calculator will tell you. Or note that $$\begin{align} 72 &= 64+8= & 2^6 + 2^3 \\ 100 &= 64 + 32 + 4 = & 2^6 + 2^5 + 2^2 \\ 60 &= 32 + 16 + 8 + 4 = & 2^5 + 2^4 + 2^3 + 2^2 \\ \end{align}$$

and note that only $2^4$ is not cancelled by the same power in another sum, while all other $2^i$ are.

As $60\oplus 16 =44$ we just need to reduce the $60$ pile to a $44$ pile (the only pile here that contains the power $2^4$) to give the opponent a value of $44 \oplus 72 \oplus 100 = 0$ making him/her lose.