Scales problem

My first thought when I read this problem is that 1 should be one of the numbers. My second thought was that in order to get all the numbers, I would have to use the difference between 2 weights to get all the numbers between 1 and 40.

I started with
a+b+c+1=40      which means that
a+b+c=39

I also thought about the lower numbers. I realized that with 1, I would either need 2 or 3. I realized that 3 gave me more options, and so I figured that one of my numbers would have to be 3, which gave me
a+b+3+1=40     which means that
a+b=36

I also can say that:
a+b+3=39
a+b+3-1=38
a+b+1=37
3+1=4
3=3
3-1=2
1=1

From here, I thought about how to make 5. Since I could only use the weights once each, I would need a larger number and subtract 3 or 1 or both from it. My first thought was that since 1 and 3 are small, it would be better to use the largest number possible, which is 9. If one of the last 2 numbers is 9, the other is 27 from a+b=36. Now I have:
27+9+3+1=40
27+9+3=39
27+9+3-1=38
27+9+1=37
27+9=36
27+9-1=35
27+9+1-3=34
27+9-3=33
27+9-3-1=32
27+3+1=31
27+3=30
27+3-1=29
27+1=28
27=27
27-1=26
27+1-3=25
27-3=24
27-3-1=23
27+3+1-9=22
27+3-9=21
27+3-9-1=20
27-9+1=19
27-9=18
27-9-1=17
27-9-3+1=16
27-9-3=15
27-9-3-1=14
9+3+1=13
9+3=12
9+3-1=11
9+1=10
9=9
9-1=8
9-3+1=7
9-3=6
9-3-1=5
3+1=4
3=3
3-1=2
1=1

I'm sure that there are more ways to do this problem, although I'm not sure of what they are. Hopefully someone in class found a different solution! I think that I don't understand this problem deeply enough to come up with an extension for it. This problem was the most challenging to me of the ones we've done so far, it took me by far the most amount of time. I would have to spend more time understanding why powers of 3 work in this case to understand how to extend it.