Shop Keeper's Dilemma

Clearly, you need a 1kg weight.

Note: 1 = ½(3¹-1).

Using that, you can use a 3kg wt with the 1kg added, or subtracted (by placing it on the opposing pan), or not at all, to weigh 2, 3, or 4 kg objects.

Note: 3 + 1 = 4 = ½(3²-1).

Using those, in combination with a 9kg wt, each added, subtracted), or not at all, to weigh anything from 5 to

9 + 3 + 1 = 13 = ½(3³-1).

Etc. With 4 wts of 1, 3, 9, and 27, you can weigh any integer from 1 to

27 + 9 + 3 + 1 = 40 = ½(3⁴-1).

And with 5 wts of 1, 3, 9, 27, and 81, you can weigh any integer from 1 to

81 + 27 + 9 + 3 + 1 = 121 = ½(3⁵-1).

So you can weigh up to 121 kg samples with those 5 wts.

the weights needed then are 1,3,9,27 & 81 kg and you can measure upto 121 kg

it has to work because the sum of this GP to n terms is [ 1 (3^n - 1)] / (3 -1) = (3^n - 1) /2 and the (n+1)th term is 1 3^n so the difference between the two is (3^n +1) / 2 which is greater by 1 than (3^n -1) / 2

thus if it works for n it will work for n+1 since it works for n = 1, it will work for succeeding values

the last weight, of course, needn't actually be 81 kg that can be even 45kg as we need to weigh only upto 85 kg

$$ This is a very famous problem and the original problem stems from the French mathematician Claude Gaspard Bachet de Méziriac (1581-1638), who solved it in his famous book Problemes plaisants et dilectables qui se font par les nombres , published in 1624. The difference is the original problem uses the limit weight of $40$ .