What's the sum of all the positive integral divisors of $540$?

My approach: I converted the number into the exponential form. And found out the integral divisors which came out to be $24$. But couldn't find the sum...Any trick?


Prime factorization of 540 is $2^2\cdot 3^3\cdot 5$. Also sum of divisors of 540 equals: $$\begin{align} \sigma\left(2^2\cdot 3^3\cdot 5\right) &=\frac {2^3-1}{2-1}\cdot\frac{3^4 - 1}{3-1}\cdot \frac{5^2 -1}{ 5-1}\\ &=7\cdot 40\cdot 6\\ &= 1680 \end{align}$$


Assuming you want to know the sum of the divisors of 540, i.e. $$ \sigma_1(540) = \sum_{d \mid 540} d $$ then you can compute this by hand by either determining all divisors of 540, or you can do this a little more cleverly.

First of all, note that $540 = 2^2 \cdot 3^3 \cdot 5$. Now, one happy fact about the function $\sigma_1(n)$ is that if $n, m$ are relatively prime, then $\sigma_1(n \cdot m) = \sigma_1(n) \cdot \sigma_1(m)$. In particular, $$ \sigma_1(540) = \sigma_1(4)\sigma_1(27)\sigma_1(5) $$ and so you only need to determine these last ones. But for prime numbers, $\sigma_1(p^k)$ is easy to compute. It is $$ \sigma_1(p^k) = 1 + p + p^2 + \cdots + p^k = \frac{p^{k+1} - 1}{p-1} $$ which should let you compute the answer now.

  • 1
    $\begingroup$ Thanks a tonne...@Simon Rose😃 $\endgroup$ – Tanishka Marrott Oct 28 '16 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.