# $\tau(a)$-function for the number of divisors

Given a number $a$ with prime decomposition $p_1^{x_1}p_2^{x_2}\cdots p_m^{x_m}$ (with $x_i \in \Bbb N$ and $p_i \neq p_j$ if $i \neq j$).

How can you proof that the number of divisors of $a$ is $\tau(a) = (x_1 + 1)(x_2+1)\cdots (x_m + 1)$?

You know that each factor $p_m^{x_m}$ has $x_m+1$ divisors: $1, p, p^2, ...p^{x_m+1}$

We are counting the divisors of $$p_1^{x_1}p_2^{x_2}\cdots p_m^{x_m}$$
For each prime factor $p_i$ we have $1+x_i$ choices where the $1$ is for the case that you do not pick that particular prime number.
Thus the total number of choices are $\tau(a) = (x_1 + 1)(x_2+1)\cdots (x_m + 1)$