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

 Messages: [ Previous | Next ]
 bob vanderbrink Posts: 1 Registered: 12/12/04
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.

Date Subject Author
11/18/97 Jonathan Pearce
11/18/97 Sam @ The NIMP Team
11/19/97 jhnieto@luz.ve
11/19/97 Sam @ The NIMP Team
11/18/97 Pat (Jedi In Training)
11/18/97 H. Peter Anvin
11/19/97 David Kastrup
11/19/97 jhnieto@luz.ve
3/3/99 bob vanderbrink
3/3/99 David Kastrup
11/20/97 Jeff Erickson
11/20/97 Brian Hutchings
1/7/98 Rosa Guia
11/10/98 James W.
10/27/03 Bill Bridges
1/4/13 Raymond C Phillips, LT USN Retired