Chameleons of three colors

Let $n_g, n_b, n_c$ be the numbers of gray, brown, and crimson chameleons respectively. Then $\{ n_g, n_b, n_c \}$ mod $3$ is $\{0,1,2\}$ . The process of chameleon-meeting changes $n_g, n_b, n_c$ mod $3$ to $n_g-1,n_b-1,n_c-1$ . So the set $\{ n_g,n_b,n_c \}$ mod $3$ stays $\{ 0,1,2 \}$ .

In other words, the numbers of chameleons of different colors are always distinct mod $3$ . So there is no way they can all be the same ( $15,15,15$ ), or all be congruent to $0$ ( $0,0,45$ ).

Nice one ... Should be level 4

Write 49 in all grey chameleons, 111 in all brown chameleons, and 50 in all crimson chameleons. Note that color change doesn't alter the total sum in modulo 3, 5, and 7. The total sum congruents 2 mod 3, 2 mod 5, and 2 mod 7. Meanwhile, the expected end result is a. Make total sum divisible by 3, 5, or 7 (as 49 is divisible by 7, 111 is divisible by 3, and 50 is divisible by 5) b. Make total sum equals 15 (49+111+50), which is divisible by all 3, 5, and 7 It's clear that such endpoints are impossible