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Math Help - Chameleon problem

  1. #1
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    Chameleon problem

    There are chameleons on an island. At a certain moment there are 13 grey, 15 brown and 17 red chameleons. If two chameleons of different colors meet they will both change to the third color. This continues in other meetings. Prove that all of the chameleons on the island never have the same color.

    Good luck!
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  2. #2
    MHF Contributor arbolis's Avatar
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    This is quite easy. I'll do it by intuition.
    At a certain moment there are 13 grey, 15 brown and 17 red chameleons
    Okay. Now say they converted like this : 44 grey, and 1 red. The red will one day meet (or if he doesn't, the problem is solved) another one who will be inevitably a grey one. They both will change to the red color. So when one is remaining in order to complete and unify the colors, he will always complicate the system of colors. Thus, there will never be all the chameleons with the same color.
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  3. #3
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    Quote Originally Posted by arbolis View Post
    This is quite easy. I'll do it by intuition. Okay. Now say they converted like this : 44 grey, and 1 red. The red will one day meet (or if he doesn't, the problem is solved) another one who will be inevitably a grey one. They both will change to the red color. So when one is remaining in order to complete and unify the colors, he will always complicate the system of colors. Thus, there will never be all the chameleons with the same color.
    But there is no logic behind the bolded sentence. I also think we need to prove that (44,1) situation will never occur.
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  4. #4
    MHF Contributor arbolis's Avatar
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    I thought about that (
    I also think we need to prove that (44,1) situation will never occur
    .). You're right, it wasn't logic. The problem is not that easy finally. Maybe if we can show that the following situation is impossible, then we're done : 1 red, 1 brown and 43 greys.
    There is an odd number of chameleons, and they always change by par (2). 3 (odd at first) subgroups that form an odd group. If 2 subgroups change colors, they will get even, but the third subgroup won't change, in order to keep the oddity of the group.
    I give a concrete example : we had 13-15-17. If the 15-subgroup meets 15 chameleons of the 17-subgroup, then the group will be 28-0-2. We see that it cannot be 1 red, 1 brown and 43 greys because of the oddity of the subgroups. It follows the group will never be 45-0-0. (Very informally and probably with error of logic.)
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  5. #5
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    Quote Originally Posted by lynch-mob View Post
    There are chameleons on an island. At a certain moment there are 13 grey, 15 brown and 17 red chameleons. If two chameleons of different colors meet they will both change to the third color. This continues in other meetings. Prove that all of the chameleons on the island never have the same color.
    If you work mod 3, the numbers 13, 15 and 17 are congruent to 1, 0 and 2. It's easy to check that if at any stage the numbers are congruent (mod 3) to 0, 1 and 2 (in some order) then the same will be true at the next stage. But 45, 0 and 0 are all congruent to 0 (mod 3), so the process could never reach that stage.
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