【Question 27】
Let $a, b, c$ be digits from 0 to 9 (repetition allowed). If all repeating decimals of the form $0 . \dot{a} \dot{c}$ are converted to their simplest fractional form, how many different values can the numerator take?
Number Theory.
$$
\begin{array}{l}
0 . \dot{a} b \dot{c}=\frac{\overline{a b c}}{999} ; \\
999=3^{3} \times 37 ;
\end{array}
$$

Consider $\overline{a b c}=001 \sim 999$;
Multiples of 3 have $999 \div 3=333$,
Multiples of 37 have $999 \div 37=27$,
Multiples of both 3 and 37 have $999 \div[3,37]=9$,
Multiples of 3 or 37 have $333+27-9=351$;
Numbers that are neither multiples of 3 nor 37 have $999-351=648$.
(1) If $\overline{a b c}$ is neither a multiple of 3 nor 37, then $\frac{\overline{a b c}}{999}$ is in its simplest form; there are 648 such numerators;
(2) If the simplest form of the fraction has a numerator that is a multiple of 37, then $\overline{a b c}$ should be a multiple of $37^{2}=1369$, which does not exist;
(3) If the simplest form of the fraction has a numerator that is a multiple of 3, then $\overline{a b c}$ should be a multiple of $3^{4}=81$; $999 \div 81=12 \cdots \cdots 27$, i.e., there are 12 multiples of $3^{4}=81$ in $1 \sim 999$; In summary, the numerator has $648+12=660$ different values.