Prove ten objects can be divided into two groups that balances each other when placed on the two pans of balance. 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.
 A: 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 $x_2>10$.
A: I don't think that there is a standard procedure, similar to the "Master Theorem", that can be applied to your problem. (Maybe someone comes up with a pigeon-hole-principle solution.) This means that we have to play with the problem by looking at cases, in order to get the feel of it. The aim is to set everything up in such a way that a minimum of cases has to be discussed.
Update
An object of weight $1$ has "underweight" $1$, an object of weight $2$ has normal weight, and any object with weight $w\geq3$ has overweight $w-2>0$, which has to be compensated by $w-2$ objects  having weight $1$.
If there is an object $A_0$ with weight $w\geq6$ then there are $\geq4$ objects $A_k$ of weight $1$. Supplement$A_0$  with $10-w\leq4$ of these $A_k$ in order to obtain $10$.
From now on we may assume that all objects have weight $\leq 5$. If there are two objects $A_1$, $A_2$ of weight $3$, $4$, or $5$, and  of combined weight $w\geq7$ then their total overweight is $\geq3$, whence there are $\geq3$ objects $A_k$ of weight $1$. Supplement $A_0\cup A_1$ with $10-w\leq3$ of these $A_k$ in order to obtain $10$.
The remaining weight patterns are of the form
$$(5,2^6,1^3), \quad (4,2^7,1^2), \quad (3^\alpha,2^{10-2\alpha}, 1^\alpha)$$
with $0\leq\alpha\leq5$, and are easy to deal with.
