Why the coin puzzle stops at twelve

Question

The famous puzzle: twelve coins, one fake that is either heavier or lighter, three weighings on a balance, identify the fake and its direction. It works.

Now try thirteen coins, with no extra genuine coin available. There are 13 × 2 = 26 possibilities and three weighings have 3 × 3 × 3 = 27 outcomes, so the simple counting argument does not rule it out.

Prove that thirteen coins nevertheless cannot be handled in three weighings.

Accepted Answer

No first weighing splits the 26 possibilities into three groups small enough for two weighings to finish. What two weighings can do. Two weighings have 9 outcomes, so after the first weighing each branch must contain at most 9 remaining possibilities. The first weighing. It must put equal numbers of coins on the pans, k on each side, leaving 13 minus 2k aside. There is no reference coin, so the pans must have equal counts or the tilt is meaningless. - If the scale balances, the fake is among the coins set aside, heavy or light: 2 × (13 minus 2k) possibilities. For this to be at most 9 we need…

Why the coin puzzle stops at twelve

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