Prove that in any set of ten different two-digit numbers one can select two disjoint subsets such that the sum of numbers in each of the subsets is the same.
My proof:
There are $2^{10} = 1024$ different subsets of the ten numbers.
Consider the possible sums of these $1024$ subsets:
Minimum sum $=$ Sum of empty set $= 0$.
Maximum sum $=$ $90 + 91 + 92 + \cdots + 99 = 945$.
There are $1024$ subsets but only $946$ possible subset sums. By the Pigeonhole Principle, there are at least two different subsets (call them A and B) such that their sums are equal.
If subsets A and B are disjoint, we have found the desired subsets.
If they are not disjoint, remove all common elements in subsets A and B (which reduces both subset sums by the same amount) to achieve the desired subsets.
Are there other ways to prove this result?
