A shopkeeper owns just four weights and a two-pan balance. Weights may be placed in either pan, so a weight in the goods pan counts as a subtraction. With his four weights he can weigh out every whole number of kilograms from 1 kg up to 40 kg. Which set of weights does he own?

Because a weight may go on either side, each weight can count as +, − or nothing, which is three choices, so the useful set is the powers of three: 1, 3, 9, 27. They add to exactly 40, and every value in between can be built, for example 2 = 3 − 1, 5 = 9 − 3 − 1 and 20 = 27 − 9 + 3 − 1. The trap is 1, 2, 4, 8: those are the right weights when everything must sit in one pan only, but they reach no higher than 1 + 2 + 4 + 8 = 15. The last two sets add to 40 as well, which is what makes them tempting, but neither can produce a value such as 2 kg or 3 kg.