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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Yet another approach. | Comment 3 of 4 |
Note the process could be run in reverse:  remove two of the same color and replace with one each of the other two.  
So we could begin with the goal 0,0,12 and see from what starting positions it can be reached. 
They are
0,0,12
1,1,10
2,2,8
0,3,9 3,3,6
4,4,4 1,4,7 1,1,10
2,5,5 2,2,8
0,6,6

The list doesn't contain 3,4,5 so that starting condition will not work. 

But Paul's solution is much more elegant. 

  Posted by Jer on 2015-07-29 18:56:37
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