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 One out of ninety (Posted on 2013-12-17)
You are presented with a set of 90 coins, appearing identical.
Actually, 89 of them are identical, but one has a different weight.
Using balance scales you have to find the odd coin,
paying \$10 each time you use the scales.

What is the maximal (i.e. the worst case) cost to accomplish it?

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 Minimum of the minimum of the maximum | Comment 2 of 7 |
I realize that I could have been more assertive in my earlier post, using an information theory argument that was within my grasp.  Even if we knew whether the odd coin was heavier or lighter, we could not guarantee isolating it in 4 weighings for anything more than 81 coins.  This is because the balance scale only has 3 different results, so 4 weighings can at best distinguish among 3^4 = 81 things.  So 5 is a minimum lower bound of the minimum of the maximum cost across all methods. Given that is achievable, is is also the actual minimum of the minimum of the maximum across all methods.

It is actually a little amazing to me that \$50 (5 weighings) is sufficient for 90 coins, given that we do not know if the odd coin is lighter or heavier.

 Posted by Steve Herman on 2013-12-17 16:27:22

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