What is the smallest number of integer weights required to exactly balance every integer between $1$ and $40$? 
What is the smallest number of integer weights required to exactly balance every integer between $1$ and $40$.

I do not really understand what this problem is asking for me to do. Any suggestions are appreciated. 

Suppose you have a set of weights consisting of specific integral weights, and you have a balance. The idea is that someone could put any integral weight up to $40$ on the left hand side, and you get to use your weights on the right hand side.
Suppose your set of weights consist of one weight each of $1,2,5,10,20$. Then you could balance some integers between $1$ and $40$, but not all of them: for example, a weight of $23$ could be balanced by putting together your $20$-weight, your $1$ weight, and your $2$-weight. On the other hand, you couldn't balance $4$, because the $1$ and $2$ aren't enough, but the $5$ is too much.
If you have enough weights, of the right sizes, to balance out any amount from $1$ to $40$, then what is the minimum number of weights you have?
The question changes if you are also allowed to add your own weights to the left side of the balance. In that case, you could balance $4$ with the above set, by putting your $1$-weight on the left, and your $5$-weight on the right. You still couldn't make $39$ though, because all of your weights only add up to $38$.
A set of "powers-of-$2$" weights is sufficient, under the first interpretation, and a set of "powers-of-$3$" weights is sufficient under the second one. The fact that the weights are meant to go up to $40$ suggests the latter interpretation, because that's the largest thing you can balance using "powers-of-$3$" weights up to $3^3$
