Question was too long to fit on title.
Fifteen different integers from 100 to 199 are given. Show that it is always possible to select from these 15 integers at least two different sets $\{a_1, b_1\}$ and $\{a_2, b_2\}$ such that the last two digits of $a_1 + b_1$ are the same as the last two digits of $a_2 + b_2$.
I know (blasted) pigeons are involved somehow, and I know in some way, if you choose 14 integers, then the 15th will overfill the "pigeonholes" and hence at least two different sets. But I don't have any idea how 14 could come out - any clues will be nice! Thanks.
Hint: Note that there are ${15 \choose 2}=105$ pairs. You are correct about pigeonholes.