Problem

Source: China Western Mathematical Olympiad 2013, problem 4

Tags: floor function, combinatorics unsolved, combinatorics



There are $n$ coins in a row, $n\geq 2$. If one of the coins is head, select an odd number of consecutive coins (or even 1 coin) with the one in head on the leftmost, and then flip all the selected coins upside down simultaneously. This is a $move$. No move is allowed if all $n$ coins are tails. Suppose $m-1$ coins are heads at the initial stage, determine if there is a way to carry out $ \lfloor\frac {2^m}{3}\rfloor $ moves