Scales Problem

If I am understanding this problem correctly in that the weights can be used on either side of the two-pan balanced (as one would naturally assume), then the solution to this problem would be to have the four masses as the first 4 powers of 3 (in grams): that is, 1g, 3g, 9g, 27g. (3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27)

I am yet to prove that this is a unique solution, but my intuition tells me so.

Reasoning/justification for solution:

First, realize that we can represent putting a weight on the same pan as the herbs as an additive inverse. If the weight is put on the opposite pan, it is considered positive.

Then, we can quickly see that using 1g and 3g weights, we can measure out 1g, 2g, 3g, and 4g of herbs.

Notice that, 4 = (9-1)/2 = (3^2 - 1)/2

That is, we can use the first two powers of 3 (starting the count at 3^0) to form a unique "scale representation" of all whole numbers up to and including (3^2 - 1)/2.

But notice that (3^k - 1)/2  = 3^(k-1) + (3^(k-1) -1)/2

Therefore, for k=3 (ie. 3 weights), we can form a unique "scale representation" of all whole numbers up to and including (3^3 -1)/2 by using weights of 3^0, 3^1, 3^2 grams.

We can iterate this reasoning to show that weights of 3^0, 3^1, 3^2, 3^3 grams (k=4) can form a unique "scale representation" of all the whole numbers up to and including (3^4 -1)/2 = (81-1)/2 = 40g

I have a formal proof by induction for any number of grams up to (3^k - 1)/2 for k weights, but it's long to share here, so I just applied the heuristic proof here. I was interested in the problem though!

This could be a good opportunity to have students think about number systems and mathematical induction.