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We have a bag of red, green and yellow apples?
Initaiily the bag contains red, green and yellow apples.
A move involves removing any two apples of different color from the bag, and replacing them with one extra apple of the third color.
After series of moves the bag finally contains one red apple.
Prove that if another series of moves was applied to the same bag, it could neever terminate in a single green or yellow apple in the end.
For example if the bag initially contained 4 apples
rrgy, we can have series of moves
rrgy
ryy
gy
r
I rather like this problem. I think the specific question posed to us is just one of a whole class of cool questions to be answered...if I have time later I might put some thought into other such questions.
Let's begin by assigning a number to each color. Almost any assignment will do, but I'll use the first one I thought of:
-1: yellow
0: red
1: green
We'll consider the sum of the numbers in the bag. For instance, in your example, our numbers are
0
-2
0
0
Consider any possible move, and how it affects our sum. If we remove RG and add Y, we remove a 1 and add a -1, for a net change of -2. Remove RY add G, we have net change of +2. Remove YG add R, and we have net change 0.
So no move can change the parity of our sum. Then, if the bag can be ended with one red apple, we have a final sum of 0, and so in each step along the way we had an even sum; in particular, we started with an even sum. Then, given any other sequence of moves, we must similarly have an even sum at each step; therefore we cannot end with a single yellow or green apple, lest our sum should be -1 or 1.
Notice that a parallel argument works with any other color, so if the bag can be reduced to a single apple of any color, it cannot be reduced to a single apple of either other color. However, it is possible to become stuck after some moves, leaving us with multiple apples all of the same color. In fact, it is possible to be so stuck as to never be able to get to a single apple. Consider, for example, starting with one apple of each color. Then any of our 3 possible moves leaves us with two apples of one color, and we are stuck. (Notice that, using the same number assignment, we're guaranteed to have an even sum in such an arrangement.)
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