The Scales problem

 There is one way that I can see to measure each integer up to 40 using only 4 weights. They would have to be weights of 1, 3, 9, and 27. The equations below calculate how we can arrive at each desired weight (1-40) using the 4 different weights (1, 3, 9, 27).

But why? Lets have a look. We need 1 to weigh 1, no way around that. If we put 1 on the other scale, the one with the items being purchased, we can use this weight of 1 as negative 1. So we don’t need 2 because we can use 3 and negative 1 to make 2. This lets us avoid having a weight of 2. Instead, we use 3 and negative 1. So we need 3. Then we can calculate 1,2,3,4 no problem using 1 and -1 and 3. Next, instead of a weight of 5, we use our weights, 1 and 3, on the other scale so negative 1 and negative 3. Therefore, we can skip 5,6,7, and 8 and include just 9 because we can get to 5 with 9 combined with both negative 1 and negative 3.

To extend this problem more deeply we could look at a general solution for calculating the next heaviest necessary weight to continue weighing things that weigh more than 40. To do this, we need to look first at the weights that we have already determined are necessary: 1,3,9 and 27. Upon first glance, I see that this could be rewritten as 3­^0, 3^1, 3^2, 3^3. Looking at it this way, we can see that the next weight will need to be 3^4. 3^4 is 81.

1,3,9, 27 are our tools that we can use to calculate each integer from 1-40.

1 = 1

2= 3-1

3=3

4=3+1

5= 9-3-1

6=9-3

7=9+1-3

8=9-1

9=9

10=9+1

11=9+3-1

12=9+3

13=9+3+1

14=27-9-3-1

15=27-9-3

16=27-9-3+1

17=27-9-1

18=27-9

19=27-9+1

20=27-9+3-1

21=27-9+3

22=27-9+3+1

23=27-3-1

24=27-3

25=27-3+1

26=27-1

27=27

…

40=27+9+3+1

 To keep going we would need 81 as our next weight.