Problem
COM-B2-M05-P006 Three Colours of Stones
There are \(20\) red, \(21\) blue, and \(22\) green stones. In one move, choose two stones of different colours, remove them, and add two stones of the third colour. Prove that it is impossible to reach a state in which all stones have one colour.
Look at differences between two colour counts modulo \(3\).
Let the numbers of red, blue, and green stones be \(r,b,g\). In a move, for example, \(r\) and \(b\) decrease by \(1\), while \(g\) increases by \(2\). The difference \(r-b\) is unchanged, while \(b-g\) and \(g-r\) change by multiples of \(3\). Therefore all pairwise differences modulo \(3\) are invariant.
Initially \(r-b=20-21=-1\), so \(r-b\equiv 2\pmod 3\). If all stones became one colour, then two of \(r,b,g\) would be \(0\), and the third would be \(63\), so all differences would be divisible by \(3\). Contradiction.
A good problem on a non-obvious modular invariant.