Who Is Left?

Centaurs, manticores, and minotaurs roam their planet. Their society is very democratic: any two animals can become sex partners. When two different species mate, their orgasm is so potent that they merge into one creature of the third species. For example, once one centaur and one manticore make love, they are reborn as one minotaur. At the beginning of the year 2016 there were 2016 centaurs, 2017 manticores, and 2018 minotaurs. They mated non-stop and at the end of the year only one creature was left on the planet. Which one?

This is one of those puzzles that I love-hate. I hate it because it is easy to answer this puzzle by inventing a specific mating pattern that ends with one animal. It is possible to get the correct answer without seeing the beauty. I love it because there is beauty in the explanation of why, if the mating ends with one animal, it has to be a specific animal.

The solution is charming, but being a mathematician this problem makes me wonder if ii is always possible to end with one animal. So I add another puzzle.

Describe the sets of parameters for which it is impossible to end up with one creature.

3 Comments

  1. none:

    A manticore as centaurs+minotaurs is even after any mating. This is probably a knee-jerk answer from a well-coached problem solver. Though it took me few minutes with matrix transformations on a paper to get the idea.
    Going by induction over the total number of creatures from (0, 1, 0) we can get any position (a, b, c) with either “both a and b even and c odd” or “both a and b odd and c even”. Actually, these patterns alternate according to the parity of a+b+c.

  2. Leon:

    It is impossible to end up with one creature if: (a) All three parameters have the same parity; or (b) Two of the parameters are zero and the other is not equal to one.

  3. Shakir:

    I find a pattern by which either 1 Manticore or 4033 Manticores would be left at the end of the gigantic exercise.

    Starting numbers:
    Ce Ma Mi
    2016 2017 2018

    1>
    If Centaurs and Manticores have the highest attraction then we would be having

    Ce Ma Mi

    0 1 4034
    1 0 4033
    0 1 4032
    1 0 4031

    0 1 2
    1 0 1
    0 1 0

    2>
    If Centaurs and Minotaurs have the highest attraction then we would be having

    Ce Ma Mi

    0 4033 2
    1 4032 1
    0 4033 0

    3>
    If Manticores and Minotaurs have the highest attraction then we would be having

    Ce Ma Mi

    4033 0 1
    4032 1 0
    4031 0 1
    4030 1 0
    4029 0 1

    3 0 1
    2 1 0
    1 0 1
    0 1 0

    If no two creatures always mate then we would not end up with a particular creature.