TITLE: Olympiad Mathematical Kosovo 2012 (Problem grade 9)
QUESTION [8 upvotes]: Let $a_{1},a_{2},a_{3},\ldots,a_{2011},a_{2012}$ be integers. Exactly 29 of them are divisible by 3. Show that $a_{1}^2+a_{2}^2+a_{3}^2+\ldots+a_{2011}^2+a_{2012}^2$ is also divisible by 3.

REPLY [4 votes]: This is very easy to solve once you know congruences, and harder when you are young and do not. That is to say a "tricky", though nice, kind of problem.
There are 2012 - 29 = 1983 squares that are all 1 mod 3, since squares can only be 0 or 1 mod 3 (if a = 2 mod 3 then a^2 = 4 mod 3 = 1 mod 3) and we have excluded all those who were divisible by 3. Their sum will be equal to 1983 mod 3 = 0 mod 3, and so divisible by 3 since 1983 is divisible by 3.