When working through this problem, I came across the solution of the weights being 1g, 3g, 9g, 27g. To demonstrate how I solved this problem, we need to think of an equal-arm weight scale like Figure 1. I will also denote the left side of the scale with a red x. When we have x+(some amount), then the items are on the same side of the equal-arm weight scale and when we have x-(some amount), then we have items on opposite sides of the equal-arm weight scale.
As the vendor has four different weights to measure herbs from 1 to 40g, the first weight must be 1g, to measure 1g herbs.
For the second weight, we need to be able to measure 2g of herbs. If the vendor has a 2g weight, then we can measure 2g of herbs, 3g of herbs, or also 1g of herbs by setting 2g on the left and 1g weight + 1g herbs on the right. If we had a 2g weight, we need a 4g weight to measure 4g of herbs. If we had a 4g weight, we need to satisfy combinations of x-1, x-2, x+1, x+2, x-1-2, x-1+2, x+1-2, x+1+2. Therefore, we can measure 3g, 2g, 5g, 6g, 1g, 5g, 3g, 7g herbs. As we can measure the same herb amounts in several ways, this may not be the most effective weight combination and we would need our last weight to be 40g-1g-2g-4g = 33g.
If our second weight
was x=3g, then we can measure x-1=2g or x+1=4g herbs. Hence now we can measure
1, 2, 3, 4 grams of herbs with our two weights.
We conclude that the third
and fourth weights must be 9g and 27g. When we have weights 1g, and 3g, we
cannot have a third weight larger than 9g as we would be unable to measure 5g
herbs at the minimum.
I believe this is the
only solution. I noticed that 1, 3, 9, 27 is part of the geometric sequence for
3y, where y is the set of whole numbers. An extension for this
activity could be to examine geometric sequences backwards in the form of
half-life, which is the time it takes for a quantity to reduce to half of its
initial value. Half-life is commonly observed in biology or nuclear physics
where substances or atoms decay.
Good work on this!
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