Example 1 Find the smallest positive integer $n$ that has exactly 144 different positive divisors, and among them, there are 10 consecutive integers.
(26th IMO Shortlist)
Analysis: According to the problem, $n$ is a multiple of the least common multiple of 10 consecutive integers, and thus must be divisible by $2, 3, \cdots, 10$. Since $8=2^{3} ; 9=3^{2}, 10=2 \times 5$, its standard factorization must contain at least the factors $2^{3} \times 3^{2} \times 5 \times 7$. Therefore, we can set
$$
n=2^{a_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdot 11^{a_{5}} \cdots \cdot p_{s}^{\alpha_{5}} .
$$

where $\alpha_{1} \geqslant 3, \alpha_{2} \geqslant 2, \alpha_{3} \geqslant 1, \alpha_{4} \geqslant 1$.
Since the number of divisors of $n$ is
$$
\begin{aligned}
d(n) & =\left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdots\left(1+\alpha_{s}\right) \\
& =144,
\end{aligned}
$$

and $\left(1+\alpha_{i}\right)\left(1+\alpha_{2}\right)\left(1+\alpha_{3}\right)\left(1+\alpha_{4}\right)$
$$
\geq(i+3)(1+2)(1+1)(1+1)=48 \text {, }
$$

there must be another $\alpha_{j}>0(j \geqslant 5)$, and $\alpha_{j} \leqslant$
2. To make $n$ as small as possible, it is advisable to take $0 \leqslant \alpha_{5} \leqslant 2$. By
$$
\begin{array}{l}
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right)\left(\alpha_{5}+1\right) \\
=144\left(\alpha_{5} \neq 0 \text { when }\right)
\end{array}
$$

or $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right)\left(\alpha_{5}+1\right)$
$$
=144 \text { (when } \alpha_{5}=0 \text {). }
$$

Considering the possible factorizations of $144=2^{4} \times 3^{2}$, and comparing the corresponding sizes of $n$, we find that the smallest $n$ is
$$
n=2^{5} \times 3^{2} \times 5 \times 7 \times 11=110880 \text {. }
$$