Ten sacks are filled with coins. In nine of them every coin weighs 10 g, but in one faulty sack every coin weighs only 9 g. A weighing machine gives an exact reading in grams, and it may be used once only. Which plan is certain to name the faulty sack?

Taking 1, 2, 3, … , 10 coins gives 55 coins, which would weigh $55\times 10=550$ g if every coin were good. Each light coin costs exactly 1 g, and the number of light coins in the pile is the number of the faulty sack, so the shortfall in grams reads off the answer directly: a reading of 543 g means sack 7. The traps: one coin from each sack always gives 99 g whichever sack is faulty, so it proves a fault exists but names nothing; splitting five against five is a two-pan idea, and in any case it only halves the field, which cannot finish in a single weighing; weighing everything and dividing by ten gives an average and cannot point at any one sack.