Solution:

For any prime $p$, let $\nu_{p}(n)$ be the maximum power of $p$ dividing $n$; i.e., $p^{\nu_{p}(n)}$ divides $n$ but not a higher power. Let $r$ be the length of the non-periodic part of the infinite decimal expansion of $1 / n$.

Write
$$\frac{1}{n} = 0 . a_{1} a_{2} \cdots a_{r} \overline{b_{1} b_{2} \cdots b_{s}}$$

We show that $r = \max \left(\nu_{2}(n), \nu_{5}(n)\right)$.

Let $a$ and $b$ be the numbers $a_{1} a_{2} \cdots a_{r}$ and $b = b_{1} b_{2} \cdots b_{s}$ respectively. (Here $a_{1}$ and $b_{1}$ can both be $0$.) Then
$$\frac{1}{n} = \frac{1}{10^{r}}\left(a + \sum_{k \geq 1} \frac{b}{\left(10^{s}\right)^{k}}\right) = \frac{1}{10^{r}}\left(a + \frac{b}{10^{s} - 1}\right)$$

Thus we get $10^{r}\left(10^{s} - 1\right) = n\left(\left(10^{s} - 1\right)a + b\right)$. It shows that $r \geq \max \left(\nu_{2}(n), \nu_{5}(n)\right)$. Suppose $r > \max \left(\nu_{2}(n), \nu_{5}(n)\right)$. Then $10$ divides $b - a$. Hence the last digits of $a$ and $b$ are equal: $a_{r} = b_{s}$. This means
$$\frac{1}{n} = 0 . a_{1} a_{2} \cdots a_{r-1} \overline{b_{s} b_{1} b_{2} \cdots b_{s-1}}$$

This contradicts the definition of $r$. Therefore $r = \max \left(\nu_{2}(n), \nu_{5}(n)\right)$.