There are A white, B black, and C red chips on a table. In one step, you may choose two chips of different colors and replace them by a chip of the third color. If just one chip will remain at the end, its color will not depend on the evolution of the game. When can this final state be reached?
Reference:Problem Solving Strategies(The book does come with solutions).
Is there a way I can solve this without looking for an invariant?
Each move changes the parity of all three colors and reduces the total number of chips by one. Since the final state has the numbers of colors disagreeing on parity the initial state must not have the colors all agree. The final color will be the odd color out in parity at the start. In fact, the total number of moves is one less than the starting number of chips. We have not proven that if you start with the piles not all the same in parity that you can get to a single chip, but that was not requested.
For fun, we can show that if the starting parities are not all the same and you start with at least two colors we can get to a single chip. Clearly if you start with only one color of chip you are stuck. You can remove two chips from any pile as long as you have two to remove and one other chip, for example $R+B\to W, R+W \to B$ will remove two red chips. We can proceed like this until all piles have two or less chips, leading to $(1,1,0) \text { or } (1,0,0)$ In the second case we are done and in the first the obvious move finishes.
