What is the smallest number of weights in a set that can be divided into 4, 5, and 6 equal-mass piles?
Let's first prove that the set cannot contain fewer than eleven weights. Suppose this is not the case, that is, there are no more than ten. Let \(60 \mathrm{~m}\) be the total mass of all weights in the set. First, note that the set cannot contain weights with a mass greater than \(10 \mathrm{~m}\) (since the set can be divided into six piles, each with a mass of \(10 \mathrm{~m}\)). Therefore, in each of the five piles with a mass of \(12 \mathrm{~m}\), there are at least two weights. Therefore, there are at least ten weights in total. This means there are exactly ten weights, and each of the five piles with a mass of \(12 \mathrm{~m}\) consists of exactly two weights. When we distribute these ten weights into six piles, there will be at least two piles with only one weight each. Therefore, there are at least two weights with a mass of \(10 \mathrm{~m}\). And thus, there are at least two weights with a mass of \(2 \mathrm{~m}\) (weights that lie in the same piles with a mass of \(12 \mathrm{~m}\) as the weights with a mass of \(10 \mathrm{~m}\)). Clearly, there cannot be four weights with a mass of \(10 \mathrm{~m}\).

Suppose there are three weights with a mass of \(10 \mathrm{~m}\). Then there are at least three weights with a mass of \(2 \mathrm{~m}\). The remaining four weights together have a mass of \(24 \mathrm{~m}\). In three piles with a mass of \(15 \mathrm{~m}\), there is one weight of \(10 \mathrm{~m}\) each, and there must be one of the remaining four (since weights of \(10 \mathrm{~m}\) and \(2 \mathrm{~m}\) cannot form \(15 \mathrm{~m}\)). This means that three of these four have a mass of no more than \(5 \mathrm{~m}\), so the last one has a mass of at least \(24 \mathrm{~m} - 15 \mathrm{~m} = 9 \mathrm{~m}\). Therefore, it has a mass of exactly \(9 \mathrm{~m}\), otherwise there would be four weights with a mass of \(10 \mathrm{~m}\). There is a pile of \(10 \mathrm{~m}\) where the weights are \(9 \mathrm{~m}\) and \(1 \mathrm{~m}\). The remaining two will have a total mass of \(14 \mathrm{~m}\), but each of them weighs no more than \(5 \mathrm{~m}\). Contradiction.

Suppose there are two weights with a mass of \(10 \mathrm{~m}\). Then four piles with a mass of \(10 \mathrm{~m}\) contain exactly two weights each. There are two weights with a mass of \(2 \mathrm{~m}\) (forming piles of \(12 \mathrm{~m}\) with weights of \(10 \mathrm{~m}\)). Therefore, there are two weights with a mass of \(8 \mathrm{~m}\) (forming piles of \(10 \mathrm{~m}\) with weights of \(2 \mathrm{~m}\)). Further, there are two weights with a mass of \(4 \mathrm{~m}\) (forming piles of \(12 \mathrm{~m}\) with weights of \(8 \mathrm{~m}\)), and there are two weights with a mass of \(6 \mathrm{~m}\). We get that all weights have a mass of \(2 k \mathrm{~m}\). But from them, it is impossible to form any pile of \(15 \mathrm{~m}\).

Contradiction

It remains to prove that there exists a set of eleven weights. For example, a set of weights with masses 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 10 grams fits. Checking that this set can be divided into 4, 5, and 6 piles of equal mass, we leave to the reader as an exercise.