1. 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 $n=5, x=37001$, we have $x_{1}=37001, x_{2}=13700, x_{3}=$ 01370 ( = 1370), $\left.x_{4}=00137(=137), x_{5}=70013\right)$.

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]
$$\text { Let } \begin{aligned}
x_{1}=x & =a_{1} a_{2} \cdots a_{n} \\
& =a_{1} 10^{n-1}+a_{2} 10^{n-2}+\cdots+a_{n-1} 10+a_{n},
\end{aligned}$$

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 $1989\left|x_{2}, 1989\right| x_{1}, 1989=9 \times 13 \times 17$,
$$a_{n} \in\{0,1,2, \cdots, 9\}$$

we know that $17\left|\left(10^{n}-1\right), 13\right|\left(10^{n}-1\right)$.
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 $17 । \left(10^{n}-1\right)$ 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 $13 । \left(10^{n}-1\right)$ 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 only need to make $x_{n}$ the smallest.
Since $1989 \mid x_{n}$, and $9 \mid x_{n}$, the smallest $x_{n}$ is
$$x_{n}=1989 \times 9=17901$$