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

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Bill Bridges

Posts: 1
Registered: 12/13/04
Re: 12 billiard ball problem
Posted: Oct 27, 2003 7:57 PM
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The original and exceedingly more difficult problem is similar to that stated below; however, one ball is slightly different weight than the other 11 balls (either heavier or lighter), but we're not sure which. The solution must a) identify the odd ball, AND, b) determine if the ball is heavier OR lighter. Again, only 3 comparison weighs using the "jeweler's scale".

Solving this problem has stumped many a genius!
h

On 18 Nov 1997, Pat Fleckenstein wrote:
><pre>
>In article <01bcf44e$fef00320$8de52ac2@jon> "Jonathan Pearce" <jvpearce@classic.msn.com> writes:

>>Does anybody know of an elegant solution to the twelve billiard ball
>>problem?
>>
>>(Twelve apparently identical billiard balls, one of which is slightly
>>heavier than the rest, determine which is the heavier one making only three
>>weighings.)

>
> Step 1:
> Put four balls on each side of the balance. If one side is
> heavier, use those balls in the next step. If neither side is
> heavier, use the four balls that are not on the balance.
>
> Step 2:
> Put two balls on each side of the balance. Use the
> heavier side in the next step.
>
> Step 3:
> Put one ball on each side of the balance. One will be
> heavier.
>
> Alternate Step 2:
> Put one ball on each side of the balance. If one side is
> heavier, you are done. If not, use the left-over balls
> in the next step.
>
> alter,
> pat
>
>--
>You have achieved excellence as a leader when people will follow you
>anywhere, if only out of curiosity.
> -- Colin L. Powell
>
>
></pre>






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