Long ago, there existed a species of fighting chameleons. These chameleons were divided into six types of matching color and strength:

Black were the strongest, followed by

blue,

green,

orange,

yellow and

white which were the weakest.

Whenever two chameleons of the same color met, they would fight to the death and the victor would become stronger and change color (eg white to yellow). Black chameleons would fight eternally.

The small island of Ula was initially populated by a group of fighting chameleons. For this group

a) the colors present each had an equal number of chameleons (for example, group = 3 black, 3 green and 3 yellow)

b) it was not made up entirely of white chameleons

After all the possible fighting was done, there remained one black and green and no blue or orange chameleons.

How many white chameleons remained in the island? Prove it.

(In reply to I must be missing a big part of the puzzle... by Erik O.)

That is what I thought of and was going to post it until I saw this.  Sure it makes this a weak problem but it is a valid solution as far as I am concerned.  I think if it stated 'fighting did occur and after all possible fighting' then of course our solution would not be valid.  I think if fighting occurred then there would be 0 white left no matter what as many have proved.  If the white population was even at the beginning, simply stated whites would either die or promote to yellow.  If there was an odd number of whites there would be one left, but that is not possible because of the end conditions.  There would have to be three of the other color represented. if 3 blk then blk at end, if 3 blu then 1 blu at end, if 3 grn then 1 blu at end.

Posted by john on 2005-06-08 17:08:14