Find the number of divisors (positive) of n (a natural integer) as a function of its prime factorization.
If $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, each positive divisor of $n$ corresponds to exactly one $d=\prod_{i=1}^{k} p_{i}^{\beta_{i}}$ with $0 \leq \beta_{i} \leq \alpha_{i}$. Each $\beta_{i}$ can therefore be chosen independently from $\alpha_{i}+1$ values. $n$ thus has $\prod_{i=1}^{k}\left(\alpha_{i}+1\right)$ positive divisors.