Question

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

Answer 1

Prime factors of $540=2^2\times 3^3\times 5$

Now, Sum of divisors $=\frac{2^3-1}{2-1}\times \frac{3^4-1}{3-1}\times \frac{5^2-1}{5-1}$ $=\frac{7\times 80\times 24}{2\times 4}=1680.$