Solution #93: A Weighty Problem II

Let us call the balls A, B, C, D, ..., L. Start by weighing four against four. If they balance, then weigh any of the remaining three against any three of the good balls. If they balance then we know the odd one is the remaining ball and we can identify whether it is heavier or lighter in the final weighing. If the three against three do not balance then we take the three containing the odd ball and weigh an one against another.

If the first weighing of four against four does not produce a balance, then the second weighing involves three against three with balls switched between the two pans and a good ball introduced. So:

	A + B + C + D > E + F + G + H

We try

	A + B + E   against   C + F + J

	A + B + E = C + F + J,

then we know that either D is heaver or G or H is lighter, so we weight G against H.

	A + B + E > C + F + J,

then we know that either F is lighter or A or B is heavier, so we weight A against B.

	A + B + E < C + F + J,

then we know that either E is lighter or C is heavier, so we weight either against a good ball (e.g. K against E).

An alternative second weighing is A + B + E against C + D + F, which follows similar lines to the above.

Source: Sloane, Paul
Categories: Reasoning, Process

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