Least number of weights required to weigh integer weights In a number theory book, I found the following problems,
"What is the least number of weights required to weigh any integral number of pounds up to 63 pounds if one is allowed to put weights in only one pan of a balance?"
and "Determine the least number of weights required to weigh any integral number of pounds up to 80 pounds if one is allowed to put weights in BOTH pans of a balance".
Now, I realize these are very simple questions, I know the answers, and I know that I can find similar questions here. 
BUT THIS QUESTION IS NOT A DUPLICATE.
I know, its easily solved by using bases 2 and 3 to write the maximum weight to be measured. What I don't understand is why? 
I must be overlooking a simple fact. 
Could someone explain me the reasoning primarily?
 A: Let's generalize the case where we are using only one pan. Say we have to weigh from 1 to n pounds(integer). We assume that a set of positive integral weights, A, exist so that we can combine them to accomplish our task. And we let |A|, the cardinality of A, be j. In other words we assume that j is the least number of weights required to weigh pounds up to n-pounds. We can combine those weights on one side of the pan in:
 $
        \begin{pmatrix}
        j \\
        1  \\
        \end{pmatrix}
$ + $
        \begin{pmatrix}
        j \\
        2  \\
        \end{pmatrix}
$ +...+$
        \begin{pmatrix}
        j \\
        j-2 \\
        \end{pmatrix}
$+ $
        \begin{pmatrix}
        j \\
        j-1  \\
        \end{pmatrix}
$ +$
        \begin{pmatrix}
        j \\
        j  \\
        \end{pmatrix}
$ ways. Where $
        \begin{pmatrix}
        j \\
        1  \\
        \end{pmatrix}
$  represents the number of ways of choosing 1 weight from j to place on one side of the pan. 
Now, since there are n integers between 1 and n we want the total number of combinations to be n, in other for them to possibly weigh n different pounds. So we want $ n=
        \begin{pmatrix}
        j \\
        1  \\
        \end{pmatrix}
$ + $
        \begin{pmatrix}
        j \\
        2  \\
        \end{pmatrix}
$ +...+$
        \begin{pmatrix}
        j \\
        j-2 \\
        \end{pmatrix}
$+ $
        \begin{pmatrix}
        j \\
        j-1  \\
        \end{pmatrix}
$ +$
        \begin{pmatrix}
        j \\
        j  \\
\end{pmatrix}
$. We realize that the terms are the coefficients of $(1+1)^j$ (binomial expansion) minus the the first coefficient 
$
        \begin{pmatrix}
        j \\
        0  \\
\end{pmatrix} 
$. So  $ n=(1+1)^j-1=
        \begin{pmatrix}
        j \\
        1  \\
        \end{pmatrix}
$ + $
        \begin{pmatrix}
        j \\
        2  \\
        \end{pmatrix}
$ +...+$
        \begin{pmatrix}
        j \\
        j-2 \\
        \end{pmatrix}
$+ $
        \begin{pmatrix}
        j \\
        j-1  \\
        \end{pmatrix}
$ +$
        \begin{pmatrix}
        j \\
        j  \\
\end{pmatrix}
$. This now implies that $n=2^j-1$. We can easily find what j is by $log(n)=log(2^j-1)$. This implies $j=log(n+1)/log(2)$. So given any integer n, if  $j=log(n+1)/log(2)$ generates an integer, then this is the least number of weights required for our task. 
Now, how to find what those integers are: We know that for an integer(pound) n we need j weights. In order for j specific integers to work for n no two different possible combinations should yield the same number. Otherwise we do not cover the whole set of integers from 1 to n since j numbers can only generate n different combinations. And also the total sum of the weights should equal n since j integers might generate n integers while they are not all within [1,n]. The total sum is the maximum so the other combinations are between 1 and n. Now we know a formula for $n=2^j-1$.  $n=2^j-1$=$2^{j-1}+2^{j-2}+...+2^0$. We see that the terms add up to n and there are j of them. So each term could potentially be a weight. How do we know that no two possible combination generates the same number? let A={$2^0$, $2^1$, ..., $2^{j-1}$}, Another requirement is that no combination of the elements of A should yield a member of A. using the formula $2^j-1$=$2^{j-1}+2^{j-2}+...+2^0$,  we see that a member of the set, A, cannot be generated by the sum of all the previous members (we think of A as ordered). And since the sum of any number of positive integers cannot generate a number less than the highest number in that sum we see that all the requirements are meant. I could be more rigorous here, but am tired. In any event, the same approach will work for the second part of the question. Remember the geometric progression. 
