The four weights have to be 1, 3, 9, and 27. I tried working it out from 1 to 9 using different numbers first, then I realized the first 3 weights could be 3 to the powers of 0, 1, and 2. Then I tried 3^3=27 and it worked out perfectly. I don't know why these 4 numbers worked, and there seemed to be a unique solution to each value, but it was so satisfying.
If it were a one-pan scale with 5 weights, base-2 numbers with exponents 0, 1, 2, 3, and 4 seem to work! 1 + 2 + 4 + 8 + 16 = 31.
Riddles like this are so fascinating, because it shows patterns in Number Theory, and the fact that there might have been actual uses of it in the ancient world. For students to understand this idea and problem, I think the process that I went through - trying out with small numbers, recognizing pattern, expanding on the pattern, then deductively explaining how it works using what they know (though I'm still stuck on the last step) - is a good approach to new problems. The uncertainty and trial-and-error process is an entry point.
I don't know much about number theory, but there is something about the base 3 coinciding with the 3 possibilities in this problem. In other words, every weight is either not used, or used once for either the same pan or the other pan (3 possibilities). If I represent any herb weight value as N,
N = 1a + 3b + 9c + 27d, with a, b, c, d belonging to the set {-1, 0, 1}
The above representation reminds me of linear algebra. But again, I still haven't figured out why!!
