TITLE: A standard and rigorous proof using the pigeonhole principle
QUESTION [7 upvotes]: Now there is a question:
For twenty two-digit numbers, we add two digits together. (like 12 and get the result 3) Prove in these twenty numbers, there must be at least 2 same results. 
Now I know I can assume the most extreme example which is adding 0,0 0,1 until 9,9, so I have at most 19 results. Therefore, the twentieth must be as same as one of the previous 19 ones.
But actually, I don't know the standard and rigorous proof of this problem. Can anyone help me with this standard proof?

REPLY [5 votes]: I don't like quoting the pigeonhole principle when it is nothing more than common sense. Like this:

There are only $19$ possible two-digit sums, because the smallest is $0+0$ and the largest is $9+9$ and there are only $19$ integers from $0$ to $18$. So obviously if you have $20$ two-digit numbers, they cannot all have different digit sums, because $20 > 19$.

More generally, if you have $c·k+1$ items and assign one of $c$ labels to each item, there is a label assigned to at least $k+1$ items. Proof: If the claim is false, then every label is assigned to at most $k$ items, so there can be at most $c·k$ items, contradicting the given number of items. Therefore the claim is true.