The following solution was sent in by Ian Wasson (Form 3), Ballymena Academy and we have included a diagram sent by Kang Hong Joo of the Chinese High School, Singapore. Well done also Sarah Meredith, class 1M, Madras College.

(1) Initial state:- | 12 green |

15 brown | |

18 yellow |

To have all green you must first balance the yellow and brown. To do this you can lose one yellow and gain two brown by having a yellow meeting a green chameleon to produce two brown chameleons.

Now you have:- | 11 green |

17 brown | |

17 yellow |

If each brown meets a yellow chameleon you will eventually end up with them all being green. Seventeen meetings will have occurred here plus the one which occurred early giving a total of eighteen meetings necessary to make all the chameleons green.

Original:

After 1st meeting:

After 18th meeting:

Note that after each meeting the numbers of all three colours have the same remainder when divided by 3. (Using the language of modulus (or clock) arithmetic they are said to be equivalent modulo 3.)

(2) The second part was explained by a student from St Peters' College, Adelaide.

Initial state:- | 13 green |

15 brown | |

17 yellow |

The only way to make all the chameleons the same colour is to have a difference between two numbers which is a multiple of three (i.e. 3, 6, 9, 12, 15, 18...) which means two numbers must have the same remainder when divided by three. With 13, 15, 17, the remainders are 1, 0, 2. No matter which two colours meet, the remainders will be 0, 2, 1. The next time two meet, the remainders will be 2, 0, 1. The next meeting, the remainders will be 1, 0, 2 and we are back to the beginning. They will always have different remainders so it is impossible for all to be the same colour because this can only happen when the numbers of two of the colours are reduced to zero, in which case they have the same remainder on division by 3.