Prove that there is at least one subset of 11 numbers whose sum is divisible by 11. I was trying to think in terms of pigeonhole but not much luck.
Edit: Any 11 natural numbers.
 A: This is a famous problem, let $a_1,a_2,\dots , a_{11}$ be your numbers.
Consider the subsets:
$\{a_1\}$
$\{a_1,a_2\}$
$\{a_1,a_2,a_3\}$
$\{a_1,a_2,a_3,a_4\}$
$\dots$
$\{a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9,a_{10},a_{11}\}$
If one of these subsets has sum that is $0\bmod 11$ we are done, otherwise two of these subsets must have the same sum $\bmod 11$. If this is the case, can you use these two sets to find a third set that has a sum of elements divisible by $11$?
A: Presumably you have a set of eleven distinct integers and wish to show that this set contains a subset whose sum is divisible by eleven regardless what numbers the set actually contained (you really should specify what type of numbers since otherwise it is ambiguous.  The set $\{1,\frac{1}{2},\frac{1}{3},\frac{1}{5},\frac{1}{7},\dots\}$ would not have any sum divisible by eleven)
Major Hint:
Let our set of integers be $\{a_1,a_2,a_3,\dots,a_{11}\}$
Define the sequence of numbers $b_n=\sum\limits_{i=1}^n a_i$ and consider the set of numbers formed (that is, we are looking at the set of numbers $\{0,a_1, (a_1+a_2), (a_1+a_2+a_3), (a_1+a_2+a_3+a_4),\dots\}$)
How can you then describe $b_i-b_j$ for $i>j\geq 0$?
If $b_i\equiv b_j\pmod{11}$ with $i>j\geq 0$, what can you say about $b_i-b_j$?
Can you guarantee that there is some pair $i,j$ such that $b_i\equiv b_j\pmod{11}$?
