There are $13$ white, $15$ black, $17$ red chips on a table. In one step, a person may choose $2$ chips of different colors and replace each one by a chip of the third color. Can all chips become the same color after some steps?
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$\begingroup$ You should really mention the source where you got this question from. It's a fairly commonly seen exercise, but your wording is almost identical to that of Arthur Engel, Problem-Solving Strategies, chapter 1, problem 6 (reproduced e.g. here and here, page 20).. $\endgroup$– Rand al'ThorJan 1 '18 at 18:19
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$\begingroup$ no, it is different to both if those problems, the first replaces 2 with 2, the second asks for a single piece remaining. a solution is to make a white chip and then make red chips until all the black and white are used up. $\endgroup$– JasenJan 2 '18 at 5:30
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No, this is impossible. Note that the pairwise differences between the three colors are 2, 2, and 4. Furthermore, the operation described—turning $(a, b, c)$ into $(a+2, b-1, c-1)$—changes the differences between any two colors by either 3 or 0. In order to obtain all chips of a single color, we need counts $(45, 0, 0)$, which has a pair with difference 0. However, it is not possible to get from either 2 or 4 to 0 in steps of 3, so this cannot be done.
