Problem
COM-B1-M08-P001 Fourteen Stones
#1
★☆☆☆☆ Level 1 of 5
There are \(14\) stones in a pile. In one move, a player may take \(1\) or \(2\) stones. Whoever takes the last stone wins. Who wins with perfect play?
The losing positions are multiples of \(3\).
With moves \(1\) or \(2\), the losing positions are \(0,3,6,9,12,\ldots\). From a multiple of \(3\), every move leaves a non-multiple, while from any other residue one can leave a multiple of \(3\). Since \(14\equiv2\pmod3\), the first player takes \(2\) stones and leaves \(12\). Then the first player complements each opponent move to \(3\) stones.
Basic residue problem.