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 re(5): Another way to answer it by Ken Haley)

You may be right, Ken, but I did have to fudge the position of 4 coins.

If I am not mistaken, the generalization is to (3^n-3)/2 coins in n weighings.  That's 39 coins in 4 weighings.

Your solution suggests to me that the general problem might succumb to an application of some combinatorial theorems, like the Marriage Theorem.

The great combinatorialist, Richard Wilson, has a solution which involves finite projective planes.  Sadly, I have forgotten his solution.

Posted by McWorter on 2005-06-10 22:10:56