# Finitely many steps to $n$-stone pile.

I have a combinatoric problem still unsolved:

$2n$ ($n$ is a positive integer) stones are divided into $3$ piles. In each step, we pick half of a pile which has even number of stones and move those stones to another pile. Prove that despite the number of stones each pile contains at the beginning, after finitely many steps, we can have a $n$-stone pile.

I tried to prove that after finitely many steps, there are 2 piles with the same number of stones by showing that $\min \{ |a-b|, |b-c|,|c-a|\} \to 0$ after finite steps, but it's too hard.

• Are you sure this is all the conditions? If we start with piles of say $(10,5,1)$, and every move consists of moving half the $10$-pile to the $5$-pile, we never see an $8$-pile. – Joffan Jul 19 '16 at 6:56
• Sorry. Each move depends on your choice: $(10,5,1) \to (5,5,6) \to (8,5,3)$ – Marry Mag Jul 19 '16 at 8:16
• Ah, so we are deliberately trying to get an $n$-pile using the move rule. That makes more sense now. – Joffan Jul 19 '16 at 8:27