Pigeonhole principle explanation Let $A$ be any set of $10$ positive integers.
Prove that there must exist at least $11$ subsets of $A$ having their element-sum with the same last $2$ digits.
(Here element-sum means sum of all elements in the set)  
Answer: Let $A$ be a set of any $10$ positive integers.
So $|A| = 10$. So, the set of all subsets of $A$ has cardinality $2^{10} = 1024$.
Now, there are $10^2 = 100$ ways for the last two digits of a number.
Let the pigeonholes be these "ways".
So, there is one pigeonhole (one number with some ending two digits) containing $\lceil \frac{1024}{10} \rceil = 11$ subsets of $A$.  
I'm unsure why we had to mention "element-sum" as I don't see where it's used in the proof.

The element sum is used indirectly, in the fact that, we only care for it's last two digits. There are less than one tenth the possible 2 digit endings, as there are subsets. Pigeonhole principle then implies that at least 11 subsets go to one of the subset sum endings.
