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Math Forum » Discussions » sci.math.* » sci.math.independent

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

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bob vanderbrink

Posts: 1
Registered: 12/12/04
billiard ball problem
Posted: Mar 3, 1999 7:57 AM
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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.






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