Nice explanation for simple puzzle There is a simple game with coins that goes as follows.
You have $x$ coins and two players who take turns. Each player can either remove one or two coins. The winner is the person who removes the last coin(s).
A little experimentation will show that the first player always wins unless $x$ is a multiple of $3$.  You can convince yourself this is true by solving for $x=1,2,3$ and then noticing that you can win if and only if at least one of $x-1$ and $x-2$ coins is a loss for the first player.  
This explanation isn't quite as elegant as I was hoping however. Is there nicer or more elegant explanation for the fact that you can always win unless $x$ is a multiple of $3$?

He who starts with a multiple of three loses because his opponent can maintain this property until the end. He who starts with a non multiple of three wins by leaving the opponent with a multiple of three.

$\color{blue}9\to8|7\to\color{blue}6\to5|4\to\color{blue}3\to2|1\to\color{blue}0$
$\color{blue}{10}\to9\to\color{blue}{8|7}\to6\to\color{blue}{5|4}\to3\to\color{blue}{2|1}\to0$
$\color{blue}{11}\to9\to\color{blue}{8|7}\to6\to\color{blue}{5|4}\to3\to\color{blue}{2|1}\to0$
