A Colorful Problem

Let $R_{n},B_{n},G_{n}$ be the respective number of red, blue and green chameleons after $n$ chameleon "meetings". If two chameleons of the same color meet then $R_{n+1} = R_{n}, B_{n+1} = B_{n}$ and $G_{n+1} = G_{n}$ . If two chameleons of different colors meet, then the values change as follows:

(i) red meets blue: $R_{n+1} = R_{n} - 1, B_{n+1} = B_{n} -1, G_{n+1} = G_{n} + 2$

(ii) red meets green: $R_{n+1} = R_{n} - 1, B_{n+1} = B_{n} + 2, G_{n+1} = G_{n} - 1$

(iii) blue meets green: $R_{n+1} = R_{n} + 2, B_{n+1} = B_{n} - 1, G_{n+1} = G_{n} - 1$

Now note that in each of these cases,

$R_{n+1} - B_{n+1} \equiv R_{n} - B_{n} \pmod{3}$

$R_{n+1} - G_{n+1} \equiv R_{n} - G_{n} \pmod{3}$

$B_{n+1} - G_{n+1} \equiv B_{n} - G_{n} \pmod{3}$

In other words, the pairwise modulo 3 differences between the numbers of chameleons of different colors is invariant, and in particular are the same as they are at the start of the process. So with $R_{0} = 15, B_{0} = 13, G_{0} = 17$ , we have that for any $n$ , $R_{n} - B_{n} \equiv 2 \pmod{3}, R_{n} - G_{n} \equiv 1 \pmod{3}$ and $B_{n} - G_{n} \equiv 2 \pmod{3}$ .

But if at some point all the chameleons were the same color, say $R = 45$ , then $B - G = 0 - 0 \equiv 0 \pmod{3}$ , which is impossible since we must always have $B_{n} - G_{n} \equiv 2 \pmod{3}$ . Similarly for $B = 45$ and $G = 45$ . Thus it is impossible for all the chameleons to simultaneously be the same color.

Let's assume only red and blue ones meet: after 13 meetings the blue ones are gone and there are 2 red left and 43 green. Then the next meeting of 2 different coloured chameleons will give a blue chameleon back. Hence: it's not possible to get 45 the same coloured chameleons.