10. The recurring decimal $0 . \dot{x} y \dot{z}$, where $x, y, z$ denote digits between 0 and 9 inclusive, is converted to a fraction in lowest term. How many different possible values may the numerator take?
(1 mark)
若把循環小數 $0 \dot{x} y \dot{z}$ （其中 $x 、 y 、 z$ 代表 0 至 9 之間的數字，包括 0 和 9）化戌最簡分數，分子有多少個不同的可能値？
10. 660
10. Note that $0 . \dot{x} y \dot{z}=\frac{\overline{x y z}}{999}=\frac{\overline{x y z}}{3^{3} \times 37}$ and $\overline{x y z}$ may range from 001 to 999 . Clearly each number from 1 to 999 that is relatively prime to 999 (i.e. not divisible by 3 and 37) gives rise to a possible value of the numerator. The number of such possible values is
$$
999-\left(\left[\frac{999}{3}\right]+\left[\frac{999}{37}\right]-\left[\frac{999}{3 \times 37}\right]\right)=648 \text {. }
$$

Now it remains to count those numbers from 1 to 999 which are not relatively prime to 999 and which give rise to new possible values of numerators. These are precisely numbers divisible by $3^{4}$ (we need not care about the factor 37 as $37^{2}$ already exceeds 1000). There are 12 such numbers, namely, $81,162, \ldots, 972$, each of which gives rise to a new possible numerator (they are $3,6, \ldots, 36$ respectively). It follows that the answer is $648+12=660$.