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.

First, give each color a level number:
L(white) = 1
L(yellow) = 2
L(orange) = 4
L(green) = 8
L(blue) = 16
L(black) = 32

Then define the level product for each color as the number of chameleons of that color times the level number. p(n) = N(n)*L(n)

Finally the Population number is P = ∑p(n) =∑N(n)*L(n)

When two chameleons of the same color meet and fight, p(n) is reduced by 2*L(n), and p(n+1) is increased by L(n+1) = 2*L(n), so P remains unchanged.

P for the ending position is 40 plus 0, 1, 2, or 3, depending on whether there were any white or yellow chameleons.

The first starting condition requires that for some number m, the number of chameleons at each level is either m or 0, so p(n) = m*L(n) or p(n) = 0. Thus P is evenly divisible by m.

Since 41 and 43 are prime, m in those cases could only be 41 or 43 (all white) which violates the second starting condition.

So P = 40 or P = 42, and there are no white chameleons left.

Posted by TomM on 2002-07-30 04:04:41