combinatorics: Repeating a procedure, will $0$ ever show up? 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.
 A: 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 $0$. 
A: When we clear two numbers and write their difference two cases could happen:
If $a$ and $b$ (the two numbers we've cleared) have the same parity then $a-b$ will be even. Otherwise, $a-b$ will be odd.
Now, from $1$ to $50$ there are exactly $25$ odd numbers and $25$ even numbers. If the first case happens, then we either will have $23$ odd numbers and $25$ even numbers, or we will have $25$ odd numbers and $23$ even numbers. In both cases after adding $a-b$ we will have an odd number of odds in our sum. So the parity of the sum will be odd. (Just look at what we will have mod $2$). If the parity is different, then we will have $24$ odd numbers and $24$ even numbers. But $a-b$ is odd in this case, therefore again the number of odds will be odd and the parity of the sum will be odd. Now apply this idea to each step and you'll see that the answer is negative because what we'll have is odd but 0 is even.
