9.1. Find the smallest four-digit number that is not a multiple of 10 and has the following property: if the digits are reversed, the resulting number is a divisor of the original number, and the quotient is different from one.

Answer. 8712.
Solution. Let's find all four-digit numbers with the property specified in the problem. Let $\overline{a b c d}$ be the desired number, then $\overline{c b a d}$ is the number obtained from it by reversing the digits, $a, d \neq 0$. According to the condition, $\overline{a b c d}=k \cdot \overline{c b a d}$, where $k$ is a natural number, $k>1$. On one hand, we have $1000 k d k \cdot \overline{d c b a}=\overline{a b c d}>1000 a$, from which $k(d+1)>a, d>\frac{a}{k}-1$. Therefore, $d=\left[\frac{a}{k}\right]$, where $[x]$ denotes the integer part of the number $x$ (i.e., the smallest integer not exceeding $x$). Then $a \geqslant k$, otherwise $d=\left[\frac{a}{k}\right]=0$.

Notice also that the last digit of the number $a k$ must be equal to $d=\left[\frac{a}{k}\right]$. Let's create a table, in the cells of which for each pair $a$ and $k$ we will sequentially write the value $\left[\frac{a}{k}\right]$ (if it is different from zero) and the last digit of the number $a k$.

| $k \backslash a$ | 2 | 3 |  |  | 4 |  | 5 |  |  | 6 |  | 7 |  |  | 8 |  |  | 9 |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 1 | 4 | 1 | 6 | 2 | 8 | 2 | 0 | 3 | 2 | 3 | 4 | 4 | 6 | 4 | 8 |  |  |  |
| 3 |  |  | 1 | 9 | 1 | 2 | 1 | 5 | 2 | 8 | 2 | 1 | 2 | 4 | 3 | 7 |  |  |  |
| 4 |  |  |  |  | 1 | 6 | 1 | 0 | 1 | 4 | 1 | 8 | 2 | 2 | 2 | 6 |  |  |  |
| 5 |  |  |  |  |  | 1 | 5 | 1 | 0 | 1 | 5 | 1 | 0 | 1 | 5 |  |  |  |  |
| 6 |  |  |  |  |  |  |  |  | 1 | 6 | 1 | 2 | 1 | 8 | 1 | 4 |  |  |  |
| 7 |  |  |  |  |  |  |  |  |  | 1 | 9 | 1 | 6 | 1 | 3 |  |  |  |  |
| 8 |  |  |  |  |  |  |  |  |  |  |  | 1 | 4 | 1 | 2 |  |  |  |  |
| 9 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | 1 | 1 |  |  |

Here, the cells are highlighted where both digits are the same. Thus, the required condition can only be met in two cases: 1) $a=8, k=4$ (then $d=2$) and 2) $a=9, k=9$ (then $d=1$).

In the first case, for the digits $b$ and $c$, we get the equation

$$
8000+100 b+10 c+2=4 \cdot(2000+100 c+10 b+8) \Leftrightarrow 2 b=13 c+1
$$

If $c=0$, then $2 b=1$, which is impossible. For $c \geqslant 2$, we get $b=\frac{13 c+1}{2}>13$, which is also impossible. The only remaining option is $c=1, b=7$.

In the second case, for the digits $b$ and $c$, we get the equation

$$
9000+100 b+10 c+1=9 \cdot(1000+100 c+10 b+9) \quad \Leftrightarrow \quad b=89 c+8
$$

For $c \geqslant 1$, we have $b=89 c+8 \geqslant 97$, which is impossible. Therefore, in this case, $c=0, b=8$.

Thus, the numbers 8712 and 9801 possess the property specified in the problem.