Let a, b, c be the count of each species in no particular order. Three operations exist: 1. a++; b--; c-- 2. a--; b++; c-- 3. a--; b--; c++

Note that a' = (b+c) either decreases by two or is unchanged. This implies two properties: a' cannot change from odd to even. a' cannot increase.

The surviving species at the end (a' = 0) has to be γ, since the others have odd initial a'. And no more than 29 can be of this type, limited by the initial a' of the α species.

@khird My thought was the following: First, the atoms perform operation (3) ten times and then operation (1) is performed 14 times. Not very likely but possible. The pot ends up with z=91-10+14=95 alpha atoms.

Sounds like you aren't implementing this line from the problem description : "When he returns, he notices that all Xmasium-atoms in the pot now are of the same type."

elyK@khird@qoto.orgSolution

@pschwede

Let a, b, c be the count of each species in no particular order. Three operations exist:

1. a++; b--; c--

2. a--; b++; c--

3. a--; b--; c++

Note that a' = (b+c) either decreases by two or is unchanged. This implies two properties:

a' cannot change from odd to even.

a' cannot increase.

The surviving species at the end (a' = 0) has to be γ, since the others have odd initial a'. And no more than 29 can be of this type, limited by the initial a' of the α species.