Please feel free to give this question a more appropriate title, as I didn't really know how to describe the problem briefly. This question concerns a exercise from A Concise Introduction to Pure Mathematics, Third Edition by Martin Liebeck,more precisely 2nd chapter, exercise 8. I will paraphrase the question:
Imagine we have a total of 30 lizards, 15 red ones, 8 green ones, 7 blue ones. When two lizards of the same color come meet, they turn into the two other colors (for example, when two red lizards meet, one turns green and the other blue). When two lizards of different colors meet, they both turn into the third color. Is it possible to have all red lizards at some instant in the future? (Hint: Consider the remainder of each color when you divide by three.)
My first instinct is to play around a little with the numbers. It seems to me like it's not possible (instinctively, there's no way to get the number of Blues to match the number of Greens). Fair enough. Next instinct is to take the hint. If we look at the remainders when we divide by three, we obtain the following configuration: $$0R - 2G - 1B$$
After some playing around, it becomes evident that this whole modulo three thing actually describes the configuration. Any operation that I do on the actual numbers I can reflect on this modulo three business. (If I perform R + R on the numbers, and then take the remainders of division by three, I obtain the same results as performing R + R as directly on the modulo three configuration.) After some more playing around, it becomes obvious that the configuration is always going to be $$2- 1 -0$$
A short proof (I hope this actually proves it). Our initial configuration is $2- 1 -0$, therefore the only possible operations are: the two same objects meet, or the one lone object meets one of the two similar objects. In both cases we end up with a $2-1-0 $ configuration.
Therefore there can never be 30 red lizards at once, since that would imply a configuration $0 - 0 -0$.
My questions are:
- is my reasoning correct?
- Why does mod 3 "encompass" the operations so well? Why mod 3?
- How can I formalize this argument? What kind of mathematics if this?