Question

Find the sum of the divisors of $18$

Answer 1

Standard form of $18$ is $2^1\times 3^2$

$\sigma \left(18\right)=\left(\frac{2^{1+1}-1}{2-1}\right)\times \left(\frac{3^{2+1}-1}{3-1}\right)$

$=\left(\frac{2^2-1}{2-1}\right)\times \left(\frac{3^3-1}{3-1}\right)$

$=\left(\frac{4-1}{2-1}\right)\times \left(\frac{27-1}{3-1}\right)$

$=\left(\frac{3}{1}\right)\times \left(\frac{26}{2}\right)$

$=3\times 13=39$

$\therefore$ Sum of divisors of $18=39$