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.

The problem states "After all the possible fighting was done, there remained one black and green and no blue or orange chameleons." And the question is how many white ones were left?
In all of the cases there will be no white chameleons left. The only exception being if there was only one white chameleon. In which case there would be no fighting in the white group. There is again no mention of number of yellow ones left. Which again will be either 1 or 0, depending on whether the white group had a fight or not. As the winner would get converted to yellow.

As for this problem there is absolutely no need for complex mathematical principles to be used. Had the question been different with different numbers in group, may be it would have been neccessary, but definitely not for above problem, right isnt'it?

Posted by Gautam on 2003-01-21 15:06:45