Question

Find the sum of all the divisors of the number 21600 ?

Answer 1

$n=21600=2^5\times 3^3\times 5^2$
$n$ has been expressed in the standard form $p_1^{k_1},p_2^{k_2}...,p_r^{k_r}$
Here, $p_1=2$, $p_2=3$, $p_3=5$
and $k_1=5$, $k_2=3$, $k_3=2$.
$\therefore d\left(n\right)=$ number of divisors of $n(=21600)$
$=(k_1+1)(k_2+1)(k_3+1)=(5+1)(3+1)(2+1)=6\times 4\times 3=72$
$S\left(n\right)=$ Sum of divisor of $n(=21600)$
$=\left(\frac{p_1^{k_1+1}-1}{p_1-1}\right)\left(\frac{p_2^{k_2+1}-1}{p_2-1}\right)\left(\frac{p_3^{k_3+1}-1}{p_3-1}\right)$
$=\left(\frac{2^6-1}{2-1}\right)\left(\frac{3^4-1}{3-1}\right)\left(\frac{5^3-1}{5-1}\right)$
$=63\times 40\times 31=78120$