Solution
Five Weights on Both Pans
Show the problem again
You have a two-pan balance and a set of exactly five weights, whose masses you may choose freely. To weigh an object, you put it on the left pan and may place any of the weights on either pan, and the object's mass is read off when the pans balance exactly. Find the largest integer N such that a set of five weights exists that can weigh every integer mass 1, 2, ..., N.
Worked solution
The answer is 121. Each weight can go on the left pan, on the right pan, or stay off, so five weights give 3^5 = 243 arrangements. The net mass on the right minus the left changes sign when the arrangement is mirrored, and the all-off arrangement gives 0, so at most (243 - 1)/2 = 121 distinct positive values are possible, meaning N ≤ 121. The weights 1, 3, 9, 27, 81 achieve this: every integer from 1 to 121 has a balanced-ternary representation using digits -1, 0, 1, and the digit pattern says which pan receives each weight (a positive digit on the opposite pan from the object, a negative digit on the same pan). A direct enumeration of all 243 combinations confirms that 1 to 121 are all obtained and 122 is not.
Source: Bachet's weights problem (1612, 'Problemes plaisants'); the ternary solution is classical. Statement written for AxiomIQ.