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 re: solution by Cheradenine)

If m=1, then the starting situation is only one chameleon per level for the levels we start with, in which case there can be no fighting. Original problem stated that there was "vicious fighting", which eliminated this possibility. However, with the changed wording, "all possible fighting" could equal 0, in which case we have to come up with 41 or 43 by adding the level numbers, but only once per level. To reach 40, we can only add 32 and 8 (black and green) plus 1 (white) for 41 and 1 + 2 (white and yellow) for 43. In both cases, no rules are broken, and therefore there could be 1 white chameleon left, provided of course that we can assume "all possible fighting" could equal none. I prefer "vicious fighting" version, however. lol

Posted by lucky on 2002-07-30 05:37:41