Simple Algebra?

$500

I think I can pay less for my weights than Victory. Using the binary numbers is certainly the most efficient if you can only add the weights but don't forget that you can use subtraction as well by putting a weight on the gold side of the scale. In that case, you can use ternary numbers and still only need one of each denomination. If we may only use one of each denomination (which isn't necessarily true but I make that restriction to show that it can be done), I would buy one of each of the following weights: 1,3,9,27,81,243,729 at a cost of $350 All weights in increments of 1 gram can be measured with these up to 1093 grams. The first few are as follows (because it is a regular system, the rest follow by inference). If a weight is shown as '-', it means that it is used on the gold side of the scale; if it is positive, it is used on the normal weight pan. 1 = +1 2 = +3 -1 3 = +3 4 = +1 +3 5 = +9 -3 -1 6 = +9 -3 7 = +9 +1 -3 8 = +9 -1 9 = +9 10 = +9 +1 ... 13 = +9 +3 +1 14 = +27 -9 -3 -1 and so on I don't doubt that, for practical purposes, weights in powers of 5 would be more useful but you would need to introduce weights in the series 2*5^n (i.e. 2, 20 ...) and this would cost more. The same would apply to powers of 4. I believe that powers of 3 gives the most cost effective solution: 7 counter weights in all at a total cost of $350.