What's the sum of all the positive integral divisors of $540$? 
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:
$$\begin{align}
\sigma\left(2^2\cdot 3^3\cdot 5\right) &=\frac {2^3-1}{2-1}\cdot\frac{3^4 - 1}{3-1}\cdot \frac{5^2 -1}{ 5-1}\\ &=7\cdot 40\cdot 6\\ &= 1680
\end{align}$$
