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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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 One out of 92! (spoiler) | Comment 3 of 7 |
Five weighings can actually find the odd coin if there are 91 identical coins and 1 odd one (92 total).

a) Use two weighings to weigh 27 vs 27 vs 27.  If they are unequal, then we we know whether the odd coin is lighter or heavier, and in which set of 27 it is.  3 more weighings isolate it.

b) If the first two weighings all balance, then the odd coin is one of 11 coins.  Take two more weighings to weigh 3 vs 3 vs 3.  If they are unequal, then we we know whether the odd coin is lighter or heavier, and in which set of 3 it is.  One more weighing isolates it.

c) If the first four weighings all balance, then the odd coin is one of the last two.  Weight one of them against one of the 90 known normal coins.  If they are unequal, then it is the odd coin, and we know whether is is lighter or heavier.  If they are equal, then the last coin is the odd one, and we do not know whether it is lighter or heavier, but we were not required to.

This method always takes 5 weighings, no more and no less.

 Posted by Steve Herman on 2013-12-18 12:58:52

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