
Re: billiard ball problem
Posted:
Mar 3, 1999 9:10 AM


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: +492347005570 Email: dak@neuroinformatik.ruhrunibochum.de Fax: +492347094209 Institut fÃÂÃÂ¼r Neuroinformatik, UniversitÃÂÃÂ¤tsstr. 150, 44780 Bochum, Germany

