TITLE: Prove ten objects can be divided into two groups that balances each other when placed on the two pans of balance.
QUESTION [5 upvotes]: There are 10 objects with total weight 20, each of the weight being a positive integer. Given that none of the weights exceed 10, prove ten objects can be divided into two groups that balances each other when placed on the two pans of balance. Hints are appreciated.
Sorry I do not know how to start this problem, so I have not shown any efforts.

REPLY [3 votes]: Let the weights be $x_1, x_2, ... x_{10}$, such that $x_1 \neq x_2$ (otherwise all the weights would be equal, which is a trivial case). We need to find a subset of it such that the sum of its elements is $10$.
Consider, $x_1, x_2, x_1+x_2, ..., x_1+x_2+x_3 + .. +x_9$. Note that if one of them is divisible by $10$, then it must be exactly $10$, as it must be smaller than $20$. Thus if one of them is divisible by $10$, we are done, if not, by pigeonhole principle, two of them have the same remainder$\mod10$. 
If we subtract these two subsets, we have one subset whose sum is divisible by $10$, but could neither be $0$ or $20$, so it must be $10$. Note that these two can't be $x_2$ and $x_1$, as it would imply $x_2>10$.