Problem
COM-B1-M08-P002 Twenty Stones
#2
★☆☆☆☆ Level 1 of 5
There are \(20\) stones. In one move, a player may take from \(1\) to \(3\) stones. The last move wins. Who wins?
Look at the residue modulo \(4\).
The losing positions are multiples of \(4\). Since \(20\) is divisible by \(4\), the first player is in a losing position. Whatever the first player takes, the second takes enough to make the two moves remove \(4\) stones. Therefore the second player makes the last move.
Good check of the losing-position idea.