1. ## Sets

$S\subset\mathbb{N}_{99}$ that contains 10 elements. Use the Pigeonhole Principle to prove that there exists two disjoint subsets of S whose elements have the same sum.

Don't know what to do.

2. Originally Posted by dwsmith
$S\subset\mathbb{N}_{99}$ that contains 10 elements. Use the Pigeonhole Principle to prove that there exists two disjoint subsets of S whose elements have the same sum.

Don't know what to do.
I suppose you have to show that there are more disjoint subsets of $S$ than there are possible different sums of the elements of those subsets. So the first question to answer is how many different disjoint subsets of $S$ there are.
Next, if you are lucky and there are a sufficiently large number of disjoint subsets of $S$, you might be able to use a very rough estimate of how many different sums there could possibly be (like, for example, $10\cdot 99$).

3. Originally Posted by Failure
I suppose you have to show that there are more disjoint subsets of $S$ than there are possible different sums of the elements of those subsets. So the first question to answer is how many different pairs of disjoint subsets of $S$ there are.
Next, if you are lucky and there are a sufficiently large number of pairs of disjoint subsets of $S$, you might be able to use a very rough estimate of how many different sums there could possibly be (like, for example, $10\cdot 99$).
To be honest, I don't know what really needs to be shown. Let alone how to go about it.