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How can we prove by induction that we need k weightings to find a lighter coin among 3 to the kth power coins?
Q. So we are given 3 to the k'th power coins and one of them is lighter (so 3 to the k'th power -1 coins are lighter).By using a 2 pan balance how can we prove by induction that 3 to the k'th power coins in which one is lighter needs k weightings ? Note: I am aware of the first case k=2 ( 9 coins needs 2 measuremnts). But I am stuck in proving k+1'th step.
Asked by Maria R - Fri Aug 17 10:39:57 2007 - - 1 Answers - 0 Comments

A. From the cases of k = 1 and 2, you see that you are weighing the coins in 3's. eg if there are 3^1 coins, you weight 2 of them and that's enough to tell which coin is lighter. If there are 3^2 coins, you weight 2 groups of 3, then that narrows it down to 3. Then you proceed as if k = 1. Similarly by induction, if we are given that 3^k coins require k weighings, the let's suppose we have 3^(k+1) coins. We divide the coins into 3 groups of 3^k each. Take two of them and perform 1 weighing. This will tell us which group has the lighter coin. Now we've narrowed it down to 3^k, which we already know requires k weighings. So increasing k by 1, increases the number of weighings by 1.
Answered by Dr D - Fri Aug 17 11:55:07 2007