7. (10 points) Let $a$, $b$, and $c$ be digits from $0-9$, not all zero and not all nine. Convert the repeating decimal $0 . \mathrm{a}^{b}{ }_{\mathrm{c}}$ into its simplest fraction form, and the numerator can vary in different cases.
660
【Analysis】According to the problem, converting $0 . a^{b} c$ into a fraction is $\frac{\overline{a b c}}{999}, a b c$ can range from 001 to 998. We need to calculate the number of multiples of 3 (excluding multiples of 81) and the number of multiples of 37, as well as the repeated multiples of both 3 and 37, and then use the principle of inclusion and exclusion to get the result.
【Solution】Solution: $\frac{\overline{\mathrm{abc}}}{999}=0 . a b c a b c \cdots$, according to the problem, $\overline{\mathrm{abc}}=1 \sim 998$, but the result needs to be simplified to the simplest fraction, so there will be repetitions after simplification. Since $999=3^{3} \times 37$, we can get:
(1) (1) $998 \div 3=332 \cdots 2$, so there are 332 multiples of 3 in $1 \sim 998$;
(2) $998 \div 3^{4}=12 \cdots 26$, so there are 12 multiples of $3^{4}=81$ in $1 \sim 998$, which means there are $332-12=320$ numbers where $\overline{a b c}$ is a multiple of 3 but not a multiple of 81. The numerators of these numbers can all be simplified to numbers that are not multiples of 3.
(2) There are $998 \div 37=26 \cdots 36$ multiples of 37 in $1 \sim 998$, which means there are 26 numbers where $\overline{a b c}$ is a multiple of 37. The numerators of these numbers can all be simplified to numbers that are not multiples of 37.
(3) There are $998 \div 111=8 \cdots 110$ multiples of both 3 and 37 in $1 \sim 998$,
which means there are 8 numbers where $\overline{\mathrm{abc}}$ is a multiple of 111. The numerators of these numbers can all be simplified to numbers that are not multiples of 111. According to the principle of inclusion and exclusion, the number of repeated numbers after simplification is $320+26-8=338$, so the numerators have $998-338=660$ different cases.
Therefore, the answer is: 660.