|
|
Re: 12 billiard ball problem
Posted:
Nov 18, 1997 3:22 PM
|
|
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
|
|
