# Simple Algebra?

By raulgc01

@raulgc01 (306)

Philippines

October 10, 2007 7:46am CST

You are in a gold buying business, and the weights of the gold being sold to you ranges from 1 gram to 1000 grams. In order to weigh the gold you have to buy counter weights, (meaning standard weights) counter weights are classified into 1 gram, 2 grams, 3 grams, 4 grams and so on up to 1200 grams (a 1 gram of gold equals the weight of a 1 gram counter weight). If each counter weight costs $50, how much will you invest so that you will be able to weigh the golds being sold to you in an efficient manner (Assuming that a fractions of a gram does not matter)?

2 responses

@victorywp (3526)

• Malaysia

15 Feb 09

1, 2, 4, 8, 16, 32, 64, 128, 256, 512
other weights can be derived from the above. cheers!

@owlwings (40010)

• Cambridge, England

19 Feb 09

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.

@victorywp (3526)

• Malaysia

19 Feb 09

great! i have never thought of it that way! through such a discussion, we can actually improve our thinking & learn much more. it's cool!