Pigeonhole Principle: Among any seven integers, there must be two whose sum or difference is divisible by $10$
Solution 1:
Hint: If you have two numbers with the same remainder when divided by $10$, then their difference is divisible by $10$, so you are done. Otherwise, consider the six pigeonholes (remainders when divided by $10$) $\{0\}, \{1, 9\}, \{2, 8\}, \{3, 7\}, \{4, 6\}, \{5\}$. Since you have seven numbers, two of them must be in the same pigeonhole. What can you conclude?