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?

Split the twelve coins and weigh six and six on each side.

Take the side that is lighter and disregard the rest.

Split this again, into two groups of three, weigh and discard the heavier side.

Weigh two coins of the three left from the light end of the previous weigh. If they balance, the remaining coin is the light one, if they do not, then the scales will tell you which is the lighter coin.

I find this simpler than having to analyse numbers.

Posted by Jennifer on 2005-06-05 11:36:10