The original puzzle What standard can we apply to 4 weights which would enable us, with a simple 2platter scale, to weigh all wholenumber weights
from 1 to 40.
The Analysis Here again, as for The Chain, we must build up the operation from the smallest number of standard weights. If we have only 1 standard weight, it must be of unit size, of weight "1", and all it can do is measure an object of weight "1". If we have 2 standard weights, the second must be of weight "3", and we can measure objects of weights "3", and "4" by placing one, or both, of our standard weights against the unknown. We can measure an object of weight "2" by placing the standard weight "3" against the unknown, and the standard weight "1" with the unknown (acting as a subtraction). If we have 3 standard weights, the third must be of weight "9", and we can measure objects of weights from "5", to "13", using the other 2 standard weights in all possible combinations of addition and subtraction. If we have 4 standard weights, the third must be of weight "27", and we can measure objects of weights from "14", to "40", using the other 3 standard weights in all possible combinations of addition and subtraction.
Mathematics The choice of a new standard weight size is evidently made by foreseeing the subtraction of the existing standard weights from the new standard : when we have standard weight "1", we subtract it from "3" to weigh "2", when we have standard weights "1" and "3", we subtract them from "9" to weigh "5", and when we have standard weights "1", "3", and "9", we subtract them from "27" to weigh "14". If "n" represnts the number of standard weights, the largest one will be of weight "3^{n1}". and the sum of the standard weights will be (3^{n}  1) / 2 . With 10 standard weights, the largest would weigh "19, 683", and the total weight would be "29, 524" !
Pedagogy You will find an application of this in the Devices Chapter.
