Problem 1380

For each $1 \leqslant j \leqslant b^{-1}$, it is clear that $(j, b)=1$. By Lemma 4, each fraction in (58) must be a pure repeating decimal when converted to a decimal, and the length of the repeating cycle is $b-1$. Let
$$\frac{1}{b}=0 . \dot{a}_{1} a_{2} \cdots \dot{a}_{b-1}$$

If $\{x\}$ denotes the fractional part of $x$, it is easy to see from (59) that for any integer $m \geqslant 0$,
$$\left\{\frac{10^{m}}{b}\right\}=\left\{\frac{10^{m+b-1}}{b}\right\}$$

Thus, for integers $m \geqslant 0$, the set of numbers
$$\left\{\frac{10^{m}}{b}\right\} \quad(m=0,1,2, \cdots)$$

contains at most $\dot{b-1}$ distinct numbers. Furthermore, since 10 is a primitive root of $b$,
$$10^{0}=1,10,10^{2}, \cdots, 10^{b-2},$$

these $b-1$ numbers are pairwise incongruent modulo $b$. Therefore, in the set (60), there are exactly $b-1$ distinct numbers, i.e.,
$$\left\{\frac{1}{b}\right\},\left\{\frac{10}{b}\right\}, \cdots,\left\{\frac{10^{b-2}}{b}\right\}$$

On one hand, we know that
$$1,10, \cdots, 10^{b-2}$$

exactly runs through the reduced residue system of $b$, $1,2, \cdots, b-1$. Thus, (61) is precisely a permutation of the following array:
$$\frac{1}{b}, \frac{2}{b}, \cdots, \frac{b-1}{b}$$

On the other hand, it is easy to see from (61) and (59) that each number in (61) is a pure repeating decimal, all with a repeating cycle length of $b-1$, and each is composed of the same $b-1$ digits in the repeating cycle of $\frac{1}{b}$, but with a different cyclic permutation. This is exactly what we need to prove.

More generally, we have the following result.