Re: math problem
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In the worst case where the two heaviest coins met at the end you will have all the non second place coins that the heaviest eliminated (3) and all the coints the second place eliminated (3) which will take 5 moves to sort through. Therefore I think this method takes 23.
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When you weigh off the four coins to find out who is second, you can eliminate one of the coins. For example:
If the four contenders are: A, B, C, and D
weighing 1 = A vs B (assume it's A)
weighing 2 = C vs D (assume it's C)
weighing 3 = winner of 1 (A) vs. winner of 2 (C) (assume it's A)
However, the coin that lost to the loser of #3 (coin D in this case) can't be the #3 coin because it's lighter than the overall winner AND at least 2 of the 2nd place contenders.
I'm pretty sure the answer is 22.
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