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Topic: 12 billiard ball problem
Replies: 15   Last Post: Jan 4, 2013 12:07 PM

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David Kastrup

Posts: 1,472
Registered: 12/7/04
Re: billiard ball problem
Posted: Mar 3, 1999 9:10 AM
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rvanderbrink@infoave.net (bob vanderbrink) writes:

> Unless I am missing something, this seems really simple. First
> weigh: separate the 12 balls into 2 groups of 6. Weigh the 2
> groups. Discard the lighter group of 6. Second weigh: Separate the
> remaining 6 balls into 2 groups of 3 balls. Weigh the 2
> groups. Discard the lighter group of 3. Third weigh: Of the 3 balls
> remaining, weigh 2 of them separately. If one is heavier than the
> other, you have the answer. If they are equal, the one that was not
> weighed is the heaviest ball. I haven't heard this problem, so I may
> be missing the rules.


You are missing the fact that it is not known whether the ball in
question is heavier or lighter than the rest.


--
David Kastrup Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut für Neuroinformatik, Universitätsstr. 150, 44780 Bochum, Germany







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