In the famous "The Odd Coin" problem you are given twelve coins, exactly one of which is lighter or heavier than the other coins. You are to determine the counterfeit coin, and whether it is lighter or heavier than the other coins, in just three weighings with a balance.

Can you solve this problem with the additional restriction that you must decide what coins go on each pan for all three weighings before any weighing takes place?

(In reply to Answer by Ken Haley)

That solution there works quite elegantly.  When I analyzed it, I looked at which numbers were ommitted in each weighing.  The only ones never ommited, 1, 3, and 5, are arranged in such a way that if all the scales tip, we know which is heavier or lighter, and whether it's heavier or lighter.  Number 2, 10, and 12 are each weighed a single time, each at different times from each other.  And all the other numbers are only ommited once each, such that again, we know which is heavier or lighter.

I wish I had thought of this solution!

Posted by Tristan on 2005-06-04 23:04:11