TITLE: combinatorics: Repeating a procedure, will $0$ ever show up?
QUESTION [6 upvotes]: On a black board we have written the numbers $1$ $2$ $...$ $50$ in a list. Each time we clear two numbers and write their difference instead. We continue this until there is only one number left. Is it possible that the number is zero?
I guess that the answer must be negative but I don't know why. The question is from a math competition for 8th graders.

REPLY [8 votes]: The difference (or absolute value of the difference) of two numbers $a$ and $b$ has the same parity as their  sum $a+b$. 
That is, $a+b$ and $a-b$ (or $|a-b|$) are always both even or both odd. This is clear from an examination of cases. 
So through the whole erase/substitute process, the parity of the sum of the numbers on the board is the same as the original parity of the sum.
But the original sum is $\frac{(50)(51)}{2}$, which is odd. Thus the sum of the numbers on the board can never become even, and in particular can never be $0$.