Let $p = 1 + a + a^2 + \cdots + a^m$ be a prime number, where $a > 1$ and $m > 2$ are integers. Prove that $\frac{10^{p-1}-1}{p}$ is a good number.

*Proof*. Since $p = 1 + a + \cdots + a^m = \frac{a^{m+1}-1}{a-1}$ is a prime number, $q = m + 1$ must be a prime number. Given $q = m + 1 > 3$, we have $q \ge 5$. Note that $p = \frac{a^q-1}{a-1}$ divides $a^q-1$ and does not divide $a-1$, so the order of $a$ modulo $p$ is exactly $q$. Therefore, $q \mid p-1$. Let $p-1 = qL$.

Consider the remainder $t$ of $10^L = 10^{(p-1)/q}$ modulo $p$, which satisfies $t^q \equiv 1 \pmod{p}$. By Lagrange's theorem, the solutions to the congruence equation $x^q - 1 \equiv 0 \pmod{p}$ are exactly $\{1, a, a^2, \cdots, a^{q-1}\}$. Thus, we can assume $t = a^r$, where $r \in D = \{0, 1, \cdots, q-1\}$. Furthermore, $10^{kL}$ modulo $p$ is $a^{\langle kr \rangle}$, where $\langle \cdot \rangle$ denotes the remainder modulo $q$, taking values in $D = \{0, 1, \cdots, q-1\}$.

Consider $N = \frac{10^{p-1}-1}{p} = \frac{10^{qL}-1}{p} = \lfloor \frac{10^{qL}-1}{p} \rfloor$ as a $p-1$ digit number, with leading zeros if necessary. Divide $N$ into segments of length $L$ from left to right (the leftmost segment may contain leading zeros), forming $q$ segments corresponding to integers $x_0, x_1, \cdots, x_{q-1}$:
$$\begin{align*} x_0 &= \lfloor \frac{10^L}{p} \rfloor = \frac{10^L - a^r}{p}, \\ x_1 &= \lfloor \frac{10^{2L}}{p} \rfloor - 10^L \cdot x_0 = \frac{10^{2L} - a^{\langle 2r \rangle}}{p} - 10^L \cdot \frac{10^L - a^r}{p} = \frac{a^r \cdot 10^L - a^{\langle 2r \rangle}}{p}, \\ \multicolumn{2}{c}{\dots \dots \dots} \\ x_k &= \frac{a^{\langle kr \rangle} \cdot 10^L - a^{\langle (k+1)r \rangle}}{p}, \quad k = 1, 2, \cdots, q-1. \end{align*}$$
If $r=0$, then $x_0 = x_1 = \cdots = x_{q-1} = \frac{10^L-1}{p}$ can be divided into two groups, each forming a geometric sequence.

If $r \in \{1, 2, \cdots, q-1\}$, then $\{\langle kr \rangle \mid k = 0, 1, \cdots, q-1\} = \{0, 1, 2, \cdots, q-1\}$. Let
$$A = \{k \in D \mid \langle kr \rangle = 0, 1, \cdots, q-r-1\}, \quad B = \{k \in D \mid \langle kr \rangle = q-r, \cdots, q-1\}.$$
When $k \in A$, $\langle (k+1)r \rangle = \langle kr \rangle + r$, and when $k \in B$, $\langle (k+1)r \rangle = \langle kr \rangle + r - q$. Thus,
$$\{x_k \mid k \in A\} = \frac{10^L - a^r}{p} \cdot \{1, a, a^2, \cdots, a^{q-r-1}\} \text{ forms a geometric sequence,}$$
$$\{x_k \mid k \in B\} = \frac{10^L - a^{r-q}}{p} \cdot \{a^{q-r}, a^{q-r+1}, \cdots, a^{q-1}\} = \frac{a^{q-r} \cdot 10^L - 1}{p} \cdot \{1, a, \cdots, a^{r-1}\}$$
also forms a geometric sequence. This completes the proof. $\square$