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Red, White, and Blue (Posted on 2015-07-29) Difficulty: 3 of 5

A bag contains 3 red, 4 white, and 5 blue balls. Two balls of different
color are removed from the bag and replaced with two balls of the third color.

Prove that no matter how many times this procedure is repeated, it is impossible for all twelve balls in the bag to be the same color.

See The Solution Submitted by Bractals    
Rating: 4.0000 (1 votes)

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Solution proof | Comment 2 of 4 |
Suppose all of the balls end up blue.

If I pick red+white and replace them with blue, the difference (white - red) is unchanged.

If I pick red+blue and replace them with two white, then the difference (white - red) increases by 3

If I pick white+blue and replace them with two red, then the difference (white - red) decreases by 3.

In all cases, the difference (white - red) stays the same mod 3. That difference is presently 1, but if they all end up blue, the final difference would be 0, which != 1 mod 3. No operations of the type allowed can produce this change.

The analogous argument applies regardless of which color we pick to be the final one. The desired final state can only be achieved if two of the three stacks start out with counts that are equal to one another mod 3.

  Posted by Paul on 2015-07-29 15:47:02
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