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The odd coin (Posted on 2002-05-01) Difficulty: 3 of 5
In a pile, there are 11 coins: 10 coins of common weight and one coin of different weight (lighter or heavier). They all look similar.

Using only a balance beam for only three times, show how you can determine the 'odd' coin.

Open problem (i cannot solve this myself): how many more coins (with the same weight as the ten) can we add to that pile so that three weighing still suffices? My conjecture is zero, though my friend guessed that adding one is possible. The best bound we can agree upon is < 2.

See The Solution Submitted by theBal    
Rating: 3.1667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Solution for 13 coins!!! | Comment 24 of 40 |
(In reply to Solution for 13 coins!!! by Morgan)

Unless I'm missing something, you gave a very nice solution, congratulations.

 


  Posted by Hugo on 2005-12-14 17:47:12
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