The 65th IMO China National Team Selection Test

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 $t^q\equiv 1(\mathrm{mod}p)$. By Lagrange's theorem, the solutions to the congruence equation $x^q-1\equiv 0(\mathrm{mod}p)$ 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 $a^{⟨kr⟩}$, where $⟨\cdot ⟩$ 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 $x_0,x_1,\cdots ,x_{q-1}$: $$\begin{array}{rl}{x}_0& =\lfloor \frac{10^L}{p}\rfloor =\frac{10^L-a^r}{p},\\ x_1& =\lfloor \frac{10^{2L}}{p}\rfloor -10^L\cdot x_0=\frac{10^{2L}-a^{⟨2r⟩}}{p}-10^L\cdot \frac{10^L-a^r}{p}=\frac{a^r\cdot 10^L-a^{⟨2r⟩}}{p},\\ \text{multicolumn}2c\dots \dots \dots & \\ x_k& =\frac{a^{⟨kr⟩}\cdot 10^L-a^{⟨(k+1)r⟩}}{p},k=1,2,\cdots ,q-1.\end{array}$$ 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 $\{⟨kr⟩\mid k=0,1,\cdots ,q-1\}=\{0,1,2,\cdots ,q-1\}$. Let $$A=\{k\in D\mid ⟨kr⟩=0,1,\cdots ,q-r-1\},B=\{k\in D\mid ⟨kr⟩=q-r,\cdots ,q-1\}.$$ When $k\in A$, $⟨(k+1)r⟩=⟨kr⟩+r$, and when $k\in B$, $⟨(k+1)r⟩=⟨kr⟩+r-q$. 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. $\square$