Competition problem — Combinatorics

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Combinatorics Difficulty 6.1 National Olympiad Prove it Singapore

Jack and Jill play the following game: Jack throws 3 dice and Jill can select some of them, possibly none, and turn each of them to the opposite side. Jill wins if the sum of the values on the dice is a multiple of 4. Can Jill always win? (Note the game is played with standard dice where the sum of the numbers on opposite sides is 7.)

Solution

Jill can always turn the dice so that the numbers are $2a$ , $2b$ , $2c$ , i.e., all even. On the other hand she can also achieve $7 - 2a$ , $7 - 2b$ , $2c$ .
Let $S = 2a + 2b + 2c$ and $T = (7 - 2a) + (7 - 2b) + 2c$ . Then both $S$ , $T$ are even and $S + T = 14 + 4c$ . Since $S + T$ is even and not a multiple of 4, one of $S$ , $T$ is a multiple of 4. Therefore Jill can always win.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.