**6th puzzle by Dr Harry Wiggins**

You are given nine coins of the same denomination and you know that one of them is counterfeit and that it is lighter than the others. You have a pan balance which means you can put any number of coins on each side and the balance will tell you which side is heavier, but not how much heavier. Explain how you can find the COUNTERFEIT COIN in exactly two weighings.

CommentsSeparate the coins into three groups, having three coins each. Compare two of these groups. If one is lighter it has the counterfeit coin. If they have the same weight, the counterfeit coin is in the group that was not weighed.

Step 2

Take the group with the counterfeit coin, and weigh two of the coins in that group. If one is lighter, it is the counterfeit coin. If they have the same weight, then the coin that was not weighed is the counterfeit coin.

Sachin Reddy Northwood School

Since the most efficient procedure for any amount of coins is to divide it into three groups of equivalent size, then divide the the one with the counterfeit coin, into three groups of equivalent size, and so on. If you have 3^n coins, then the least number of weighings needed to find the counterfeit coin is n. For an integer k, such that 3^(a)<k<3^(a+1), the least number of weighings, is a+1. Thus for any n coins, the least number of weighings needed to determine the counterfeit coin is equal to the exponent, of the smallest power of three greater than or equal to n, or ceiling(log3(n)).

Sachin Reddy Northwood School.