How many positive divisors are there of the number $2019^{2019}$?

Question

How many positive divisors are there of the number $2019^{2019}$ ?

Since $2019$ has $4$ positive divisors $1,~3,~673,~2019$, the positive divisors of $2019^{2019}$ are

$1, \\ 3,~3^2,~3^3, \cdots, 3^{2019}, \\ 673,~673^2, ~ 673^{3},\cdots, 673^{673}, \\ 2019,~2019^2,~2019^3, \cdots, 2019^{2019}. $

So there are Total $1+3 \times 2019=2058$ positive divisors of $2019^{2019}$ according to me.

Am I right ?

Or something wrong?

Answer 1

You left out $3^2\times 673$ et al.

There are $2020\times2020$ numbers of the form $3^a673^b$ with $a$ and $b$ integers between $0$ and $2019$, and they are the factors of $2019^{2019}$.

Answer 2

If $d(n)$ denotes the number of positive divisors of $n=p_1^{e_1} \cdots p_k^{e_k}$, then

$$ d(n) = (e_1+1) \cdots (e_k+1). $$

This is because $m \mid n$ if and only if $m=p_1^{f_1} \cdots p_k^{f_k}$, where $f_i \in \{0,1,2,\ldots,e_i\}$ for each $i$, leading to $e_i+1$ choices for $f_i$.

Therefore,

$$ d(2019^{2019}) = d(3^{2019} \cdot 673^{2019}) = (2019+1)^2 = 2020^2 = 4080400. \quad \blacksquare $$