Well explained solutions to this problem came from Anna of West Flegg Middle School, Avishek, Mark, Thomas, James and Ricardo of Simon Langton Boys' Grammar School, and Nodoey from Singapore.
I start with three marbles, 1 red, 1 green and 1 blue. I can trade any one marble for two others, one each of the other two colours. However many times I do this it is impossible to have a difference of 5 between the number of red and blue marbles because the difference between these two numbers is always even.
Suppose I trade one marble of any colour (say blue) for two others, then I will have two reds and two greens and no blues. If I then trade one of my marbles I will have one blue, one of another colour and three of the third colour. Each time I trade I receive one more marble in total but, more importantly I have alternately an even number of each colour then an odd number of each colour, then an even number, and so on. For example:
R B G 1 1 1 2 0 2 1 1 3 0 2 4 1 3 3
So the number of blue and the number of red marbles are always both even or both odd (this applies to every colour). So the difference between the number of reds and blues is always even and can never be five.
A further challenge:
You might like to improve on this solution by using vectors in the proof. You can learn about adding vectors by reading the article A Knight's Journey and apply the ideas to the Marbles problem.)