
billiard ball problem
Posted:
Mar 3, 1999 7:57 AM


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.

