Set of $52$ integers, Dirichlet's principle

Question

Prove that in any set of $52$ integers we can find at least $2$ numbers, summation or subtraction of which will be divided by $100$.

Answer 1

Make the ''boxes''

$\{0,100\}$; $\{1,99\}$; $\{2,98\}$;... $\{49,51\}$ and $\{50\}$

Then put a number in box $\{i,100-i\}$ if it has a remainder $i$ or $100-i$ when divided by $100$.

Then in one box there are at least two numbers (since we have 51 boxes) and thus conclusion.

Answer 2

Hint. The fact that we have a "$52$" floating around in the question suggests that we should be looking for a set of $51$ equivalence classes, such that any two integers falling into the same equivalence class will satisfy the requirement. Since there are $52$ integers in all, some pair of integers that are satisfactory must exist.

Thus, determining what those equivalence classes are if your task. For instance, one equivalence class must contain both $8$ and $308$, since their difference is $300$, which is divisible by $100$. But just relying on equivalence modulo $100$ won't work; that requires $100$ equivalence classes. What can you do to cut the number of classes down to $51$?