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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 remains at the end, its color will not depend on the evolution of the game. When can this final state be reached?

This the solution given in the book:

Solution: All three numbers $a$, $b$, $c$ change their parity in one step. If one of the numbers has different parity from the other two, it will retain this property to the end. This will be the one that remains.

I don't understand this given solution.

Xoque55
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Bluey
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  • Do you understand what is meant by the parity of an integer? – Brian M. Scott Jun 18 '15 at 18:17
  • whether number is even or odd – Bluey Jun 18 '15 at 18:19
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    Right. Now check that at each move, the parities of the numbers of chips of each color change. For instance, if you take a white and a black chip, you lose one of each of those colors and gain one red chip, so the parities of all three numbers change. Now suppose that two of $a,b$ and $c$ are even and the third is odd, or two are odd and the third is even: one pile has a different parity from the other two. At each move you change the parity of every pile, so the pile that had a different parity to begin with still has a different parity. This must continue until the game is over. ... – Brian M. Scott Jun 18 '15 at 18:23
  • ... Now when there’s only one chip left, the pile that contains it has odd parity, and the two empty piles have even parity. Thus, the pile with the remaining chip must be the one whose parity was different from the other two all the way through the game. – Brian M. Scott Jun 18 '15 at 18:25
  • You’re welcome. – Brian M. Scott Jun 18 '15 at 18:30

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