I'm reading a book about combinatorics. Even though the book is about combinatorics there is a problem in the book that I can think of no solutions to it except by using number theory.
Problem: Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$?
My proof is kinda simple. Let's work in mod $2$. We'll have:
$\pm 1 \pm 2 \pm \cdots \pm 100 \equiv 101 \mod 2$ but since $+1 \equiv -1 \mod 2$ and there are exactly $50$ odd numbers and $50$ even numbers from $1$ to $100$ we can write:
$(1 + 0 + \cdots + 1 + 0 \equiv 50\times 1 \equiv 0) \not\equiv (101\equiv 1) \mod 2$ which is contradictory.
Therefore, it's not possible to choose $+$ or $-$ signs in any way to make them equal.
Now is there a combinatorial proof of that fact except what I have in mind?