Anse St. Louis
9/11/13
Mr.Lass, 8L
POW

Chameleons 1)We started out with twelve red fifteen blue and eighteen yellow. We had all the blue and yellow meet, so that would now make thirty new red chameleons. since there were fifteen meetings each is one pair of chameleons so we multiplied two by fifteen and got thirty and added that to the twelve red chameleons that had already existed (Forty two now in total).
Now we had forty two red no blue and three yellow. So we could get some blue chameleons back we had one red chameleon meet a yellow chameleon and the result of that would be two new blue chameleons forty one red chameleons and 2 yellow chameleons. Now to make all chameleons red we had both blue and yellow meet to make 45 red chameleons.
2) No, on problem number two we had tried guessing and checking and they wouldn’t ever be same color so we arrived at the answer that it would be impossible. We tried working out the problem a few more times to make sure but remained at the same answer. But recently we found that for all the chameleons to be the same color the two colors that must change need to be a multiple of three. Example: If all chameleos needed to be red, 13 red 15 blue and
17 yellow wouldn’t work because 17 isn’t a multiple of three but 13 red 15 blue and 18 yellow would work because both 15 and 18