Proving that for an even $k$ you can't move $k$ rocks from $k$ platforms to the same platform, given a set of rules

Question

We have $k$ rocks on $k$ different platforms placed in a circle. You can only move two different rocks at a time (not more, not less), and you can move them only one platform to the left or to the right with the following rule: after choosing the two rocks, if you move one clockwise, the other must move counter-clockwise.

If $k$ is odd, it's easy enough to see you can move them in such way that they all end up on the same platform, but I'm having trouble proving that for even $k$ values this is impossible that all rocks end up on the same platform. I would like some help proving that.

Answer 1

Number the platforms $0, 1, \ldots, k-1$. Given a configuration of rocks on platforms, for each rock take the number of its platform, and add all those numbers together, modulo $k$. Now note the following:

• In the initial configuration, this number is equal to $k(k-1)/2$, which is not zero modulo $k$; if $2^n$ is the largest power of two dividing $k$, then $2^{n-1}$ is the largest power of two dividing $k(k-1)/2$.
• Moving one rock a space clockwise and the other a space counterclockwise does not change the number modulo $k$.
• If all rocks are on the same platform, the number is $k$ times that platform number, i.e. zero modulo $k$.

Thus moving them all to the same platform using the move you describe is impossible.