Saudi Arabia Mathematical Competitions 2012

The operation is invertible; its inverse is to take consecutive counters $Y,Z,X$ with $X$ black, move $X$ to the front, and flip $Y$. We will work with this operation instead, and show that from any initial position with at least one black counter, we can reach the configuration with one black counter on position $1$.

Let $B$ denote a black counter and $W$ denote a white counter. Suppose we do not have two consecutive counters of the same color anywhere. Take a configuration $BWB$ and perform the move changing it to $BWW$. Given that we have two consecutive white counters, we can find a configuration $WWB$ and perform the move changing it to $BBW$. So no matter what we start with, we will be able to obtain two consecutive black counters somewhere.

Now we use our two consecutive black counters to turn everything black. Suppose there is a white counter; by taking the first white counter preceding a run of at least two consecutive black counters we have a configuration of the form $WBB$. Perform the move to obtain $BBB$, eliminating this white counter. By repeating this process we change all the counters to black.

It now suffices to show that we can get to just one black counter from this, since we can cyclically shift any moves from this point to ensure it ends up on space $1$. Given the configuration $BBB$, two moves will change it to $BWW$. So we can change the black counters to white in pairs. Repeat this process in a way that keeps all the white counters and black counters consecutive. If $n$ is odd, this gives us one black counter at the end. If $n$ is even, we get two consecutive black counters with the rest white. Then perform three moves on $WBB$, which changes it to $BWW$. This completes the proof.