In article <01bcf44e$fef00320$8de52ac2@jon> "Jonathan Pearce" <firstname.lastname@example.org> 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.
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