# Using your Head is Permitted

## June 2010 solution

The answer is that this is possible iff *n* is odd.
Let *Z*_{i} be the permutation mapping each *x* to
*x*+*i* mod *n*.

For an odd *n*, *A*=*B*=*Z*_{1},
*C*=*Z*_{2} is a valid solution.

For even *n*, consider the sign of permutations *A*, *B* and
*C*. All cyclic permutations can be described as a name-change over any
other cyclic permutation. So, for example, for permutation *A* there is
some one-to-one function *P*_{A}, such that
*A*=*P*_{A}^{-1}*Z*_{1}*P*_{A}.
Regardless of what the sign of *P*_{A} is, the sign of
*A* must equal the sign of *Z*_{1}, which is -1. More
generally, the sign of all cyclic permutations, including all of
*A*, *B* and *C*, must be -1.

However, the sign of *B*ο*A* is -1*-1=1, so it cannot be
a cyclic permutation.

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