All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
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.)
Some Thoughts re: Yet another approach... @Paul & Bractals Comment 4 of 4 |
(In reply to Yet another approach. by Jer)

Like Jer, I've applied the "reverse engineering" approach, trying to define what common feature is shared by the "reachable" triplets as opposed to "unreachable" ones.

Helas - I did not consider "going mod 3."

Comparing the 3 approaches presented so far I feel that to say  that Paul's solution is much more elegant is an understatement. 

It is the only general solution , valid for 3, 4, 5 triplet as well as for 333, 344,  338. (Just imagine finding all possible paths for the new triplet!) 

Excellent proof, Paul!

Nice problem (I've rated it 4), Bractals!

  Posted by Ady TZIDON on 2015-07-30 01:59:48
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information