# How do you find the number of factors of a number?

If the prime factorization of n∈N is

$n = \setminus {\prod}_{k = 1} {p}_{k}^{{a}_{k}}$

then the number of divisors of $n$ is

$\setminus {\prod}_{k = 1} \left({a}_{k} + 1\right)$

For example $36 = {3}^{2} \cdot {2}^{2}$ the number of divisors is

$\left(2 + 1\right) \cdot \left(2 + 1\right) = 9$

which are

$1 | 2 | 3 | 4 | 6 | 9 | 12 | 18 | 36$