Express the following in the form $\frac{p}{q},$ where $p$ and $q$ are integers and $q \neq 0.$
$0.\overline{125}$

(N/A) Let $x = 0.\overline{125}$.
$\therefore x = 0.125125\ldots$ $(1)$
Since there are three repeating digits,multiply both sides by $1000$.
$1000x = 125.125125\ldots$ $(2)$
Subtract equation $(1)$ from equation $(2)$:
$1000x - x = 125.125125\ldots - 0.125125\ldots$
$999x = 125$
$\therefore x = \frac{125}{999}$
Thus,$0.\overline{125} = \frac{125}{999}$.