There are 2004 girls sitting around a table. One of them has 2004 candies, and the others have none. On each round, every girl having more than 1 candy gives 1 candy to each girl to the left and right of her. Prove that the condition in which each girl has exactly 1 candy is impossible.
I can't even get any idea for this... Can anyone give me hints? Thanks a lot!
It's a very old problem, dating back to the year 1994, when it was asked for $1994$ girls. That one was really easy (mod $4$), because $1994$ is congruent to $2$ mod $4$.
For general even number $n$ (e.g. $n = 2004$ here), an argument mod $n$ suffices.
The interesting thing is that, if we start with an odd number of girls, then the procedure will eventually end in finitely many steps. The proof is more complicated and much more interesting than the original problem.
For reference: IMO 1994 shortlist Problem C5. Also see the remark there.