Help I'm stuck with z=95! https://www.mathekalender.de/index.php?page=problem&problemID=44&lang=en
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.
Where have I misunderstood the problem?
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."
@khird Made an computation error.
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