The Scales Problem

The first step I took in approaching this problem is to figure out the mass of the smallest weight, which is one gram because it's required to measure exactly one gram of herb.

Next, I proceed onto the second smallest weight (let's denote its mass as X). This weight can measure X amount of herb, as well as X+1 and X-1 amounts when used in combination with the one-gram weight. I then deduced that the smallest integer X can take is 3, because at X=3, it is possible to measure both two and four grams of herb, whereas if X=2, it can also be used to measure one gram, which is counterproductive.

Using this method, I can also find the second largest weight (let's denote its mass as Y). In a similar fashion as above, this weight can measure Y, Y+3, Y-3, Y+3+1, Y-3-1, Y+3-1, Y+1-3 grams of herb. The smallest amount of these is Y-3-1, which means Y must be 9 (since the small weights can already measure 1,2,3 and 4 grams, Y must be a number that can be used to measure at least 5 grams). Accordingly, the amounts of herb that can be measured using the three weights are: 5, 6, 7, 8, 9, 10, 11, 12 and 13 grams.

Lastly, to find the largest weight, we apply the same rules (let its mass be Z): the four weights used together must be able to measure the remaining amounts of herb (from 14 to 40). Since we know that the smallest amount we need is 14, then Z-9-3-1=14 must hold. Thus, solving the equation gives us Z=27.

So, the four weights are: 1, 3, 9, 27 grams

To extend this problem in the classroom, we can apply the geometric sequence pattern and explore how it may be related to simple addition and subtraction. This can be done by changing the parameters such as stretching the range of the maximum amount of herb, as well as allowing for multiple quantities of the same weight.