What is the minimum number of weights needed to be able to weigh any number of grams from 1 to 100 on a balance scale, if the weights can be placed on either pan of the scale?

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When solving this problem, we need the following interesting property of the ternary numeral system: any natural number can be represented as the difference of two numbers, the ternary representation of which contains only 0 and 1.

To prove this, we need to write the original number in ternary notation and construct the required numbers digit by digit from right to left. If the resulting numbers have ones in the same positions, they can be replaced with zeros.

Now it is clear that it is sufficient to have 5 weights with masses of 1, 3, 9, 27, 81 (think about why a weight of 243 grams is not needed).

Four weights, however, are clearly insufficient, as they can measure no more than \(3^4-1=80\) different weights (each weight is either on the left pan, or on the right pan, or not involved in the weighing).

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