How many divisors does 24 have? and 140? and $2012^{2013}$?
Let's directly pose the more general question of the number of divisors of the integer $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$.

For an integer $d$ to divide $n$, it is necessary and sufficient that for all $i$, the exponent of $p_{i}$ in its prime factorization is less than or equal to the corresponding exponent in the factorization of $n$. (For more information on this, refer to the paragraph on p-adic valuations in the arithmetic course notes)

Thus, if we write $d=\prod_{i=1}^{k} p_{i}^{\delta_{i}}$, we must have $\delta_{i} \in\left\{0,1, \ldots, \alpha_{i}\right\}$, which gives $\alpha_{i}+1$ possibilities for the exponent of $p_{i}$ in $d$.

Finally, the number of divisors of $n$ is: $\prod_{i=1}^{k}\left(\alpha_{i}+1\right)$

In conclusion, $24=2^{3} \times 3$ has $4 \times 2=8$ divisors, $140=2^{2} \times 5 \times 7$ has $3 \times 2 \times 2=12$ divisors, and $2012^{2013}=2^{4026} \times 503^{2013}$ has $4027 \times 2014=8110378$ divisors.