$1 \cdot 190$ Let $x=a_{1} a_{2} \cdots a_{n}$ be an $n$-digit number in decimal notation, where $a_{1}, a_{2}, \cdots, a_{n}\left(a_{1} \neq 0\right)$ are non-negative integers not exceeding 9. The following $n$ numbers are obtained by cyclically permuting $x$:
$$
\begin{array}{l}
x_{1}=x=a_{1} a_{2} \cdots a_{n}, \\
x_{2}=a_{n} a_{1} a_{2} \cdots a_{n-1}, \\
x_{3}=a_{n-1} a_{n} a_{1} a_{2} \cdots a_{n-2}, \\
\cdots \cdots \\
x_{n}=a_{2} a_{3} \cdots a_{n} a_{1} .
\end{array}
$$
(For example, when $n=5, x=37001$, then $x_{1}=37001, x_{2}=13700, x_{3}=$ $\left.01370(=1370), x_{4}=00137(=137), x_{5}=70013\right)$.

Find the smallest number $n$ such that the $n$ numbers obtained by cyclically permuting $x=a_{1} a_{2} \cdots a_{n}$ are all divisible by 1989, and prove it. Then find the smallest $x$ that satisfies the above condition.
[Solution] Let $x_{1}=x=a_{1} a_{2} \cdots a_{n}$
$$
=a_{1} 10^{n-1}+a_{2} 10^{n-2}+\cdots+a_{n-1} 10+a_{n},
$$

then $x_{2}=a_{n} a_{1} a_{2} \cdots a_{n-1}$,
thus we have $10 x_{2}-x_{1}=\left(10^{n}-1\right) a_{n}$.
Given that 1989 divides $x_{2}$, 1989 divides $x_{1}$, and 1989 = 9 × 13 × 17,
$$
a_{n} \in\{0,1,2, \cdots, 9\},
$$

it follows that 17 divides $\left(10^{n}-1\right)$, and 13 divides $\left(10^{n}-1\right)$.
Since $10^{2} \equiv 15,10^{3} \equiv 14,10^{4} \equiv 4,10^{5} \equiv 6,10^{6} \equiv 9,10^{7} \equiv 5$,
$$
\begin{array}{l}
10^{8} \equiv 16,10^{9} \equiv 7,10^{10} \equiv 2,10^{11} \equiv 3,10^{12} \equiv 13,10^{13} \equiv 11,10^{14} \equiv 8, \\
10^{15} \equiv 12,10^{16} \equiv 1(\bmod 17)
\end{array}
$$

Therefore, from 17 divides $\left(10^{n}-1\right)$, we know that $n$ is a multiple of 16.
Also, since $10^{2} \equiv 9,10^{3} \equiv 12,10^{4} \equiv 3,10^{5} \equiv 4,10^{6} \equiv 1(\bmod 13)$, from 13 divides $\left(10^{n}-1\right)$, we know that $n$ is a multiple of 6.
Thus, $n$ is a multiple of 48. The smallest value of $n$ is 48.
Furthermore, as long as 9 divides $x$, it can divide any number obtained by permuting $x$. To find the smallest number $x$, we must have $a_{1}=1$, and take as many 0s as possible from $a_{2}$ onwards.
Therefore, $x_{n}=a_{2} a_{3} \cdots a_{n} 1$, so to make $x$ the smallest, we need to make $x_{n}$ the smallest.
Since $1989 \mid x_{n}$ and $9 \mid x_{n}$, the smallest $x_{n}$ is
$$
x_{n}=1989 \times 9=17901 \text {. }
$$