Question

What is the sum of all positive integral divisors of the numbers 540?

Answer 1

Hint: In this question, we are asked to find out the sum of all positive integral divisors of a given number. As the positive integral divisors would be the numbers which are products of one or more factors of the given number, we should first find out the factors of 540. Then, as we have to find out the sum of all the integral divisors and we know that each divisor will be a product of the factors, we can write it as $\left( {{a}^{0}}+{{a}^{1}}+...+{{a}^{{{r}_{1}}}} \right)\left( {{b}^{0}}+{{b}^{1}}+...+{{b}^{{{r}_{2}}}} \right)\left( {{c}^{0}}+{{c}^{1}}+...+{{c}^{{{r}_{3}}}} \right)...$ where a,b,c etc. are the factors of 540 and the highest power of a in the factorization of 540 is ${{r}_{1}}$ , highest power of b in the factorization of 540 is ${{r}_{2}}$ and so on.

Complete step-by-step answer:
We are asked to find the sum of the positive integral divisors of 540, therefore, we should first find out the factors of 540. We can do it using the prime factorization method in which we divide a number by its lowest possible prime factor and then again divide the obtained number by the lowest possible prime factor to obtain
\[2\overline{\left){\begin{align}
  & 540 \\
 & 2\overline{\left){\begin{align}
  & 270 \\
 & 3\overline{\left){\begin{align}
  & 135 \\
 & 3\overline{\left){\begin{align}
  & 45 \\
 & 3\overline{\left){\begin{align}
  & 15 \\
 & 5\overline{\left){\begin{align}
  & 5 \\
 & 1 \\
\end{align}}\right.} \\
\end{align}}\right.} \\
\end{align}}\right.} \\
\end{align}}\right.} \\
\end{align}}\right.} \\
\end{align}}\right.}\]
Therefore, we can write 540 as $540=2\times 2\times 3\times 3\times 3\times 5\times 1={{2}^{2}}\times {{3}^{3}}\times {{5}^{1}}...............(1.1)$
Now, if a number is an integral divisor of 540, the factors in it should be such that when 540 is divided by that number, the factors cancel out. Therefore, any integral divisor of 540 should be of the form
${{2}^{p}}{{3}^{q}}{{5}^{r}}$ where $0\le p\le 2,0\le q\le 3,0\le r\le 1\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left( 1.2 \right)$
Therefore, we can write the sum of the integral divisors as
$\left( {{2}^{0}}+{{2}^{1}}+{{2}^{2}} \right)\times \left( {{3}^{0}}+{{3}^{1}}+{{3}^{2}}+{{3}^{3}} \right)\times \left( {{5}^{0}}+{{5}^{1}} \right)..................(1.3)$
As we can see that by expansion, all the terms of the form (1.2) are added in equation (1.3). Therefore, the value of the expression in (1.3) should give us the required sum of all the integral divisors of 540.
The value of the expression in (1.3) is
$\begin{align}
  & \left( {{2}^{0}}+{{2}^{1}}+{{2}^{2}} \right)\times \left( {{3}^{0}}+{{3}^{1}}+{{3}^{2}}+{{3}^{3}} \right)\times \left( {{5}^{0}}+{{5}^{1}} \right) \\
 & =\left( 1+2+4 \right)\times \left( 1+3+9+27 \right)\times \left( 1+5 \right) \\
 & =7\times 40\times 6=1680 \\
\end{align}$
Thus, the required answer to this question is 1680.

Note: We should note that we should be careful to include the terms involving zeroth power of the factors in equation (1.3). It is because a product of other two factors will also be an integral divisor of 540, for example ${{2}^{0}}\times {{3}^{2}}\times {{5}^{1}}=45$ involves zeroth power of 2 and still is a factor of 540.