Weighing

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n weighings are sufficient to find a bad coin among (3^n-1)/2 coins (Steinhaus 1999, p. 61). vos Savant (1993) gives an algorithm for finding a bad ball among 12 balls in three weighings (which, in addition, determines if the bad ball is heavier or lighter than the other 11), and Steinhaus (1999, pp. 58-61) gives an algorithm for 13 balls.

Bachet's weights problem asks for the minimum number of weights (which can be placed in either pan of a two-arm balance) required to weigh any integral number of pounds from 1 to 40 (Steinhaus 1999, p. 52). The solution is 1, 3, 9, and 27: 1, 2=-1+3, 3, 4=1+3, 5=-1-3+9, 6=-3+9, 7=1-3+9, 8=-1+9, 9, 10=1+9, 11=-1+3+9, 12=3+9, 13=1+3+9, 14=-1-3-9+27, 15=-3-9+27, 16=1-3-9+27, 17=-1-9+27, and so on.

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