Re: math problem
<font color="white">First figure out the heaviest one just doing a binary search: 15 weighings. The key point is that the second heaviest coin is one that was eliminated by the winner at some point; there are only four choices (the coins which lost directly to the ultimate winner; we also have a pool of 10 coins which contains the second and third place guys). So two more weighings can determine the second heaviest. Now do the same thing for third place...in the worst case, the second place coin from the original bracket wins. Now there are 5 contenders for third place (two from the second place pool and three that were eliminated in the original pool). This requires four more weighings. So 21 or better. It looks like 17 is simplest for top two; can one do better than 21 for the top three?</font>
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