6. Find all four-digit numbers $\overline{a b c d}$ (where $a, b, c, d$ are the digits of the decimal representation), each of which is a divisor of at least one of the three four-digit numbers $\overline{b c d a}, \overline{c d a b}, \overline{d a b c}$ formed from it.
Answer: All numbers of the form $\overline{a b a b}$, where $a$ and $b$ are any digits except zero (there are 81 such numbers).

## Solution.

From the problem statement, it follows that there exists $k \in \mathbb{N}$ such that at least one of the following equalities holds:
(I) $k \cdot \overline{a b c d}=\overline{b c d a}$
(II) $k \cdot \overline{a b c d}=\overline{c d a b}$
(III) $k \cdot \overline{a b c d}=\overline{d a b c}$.

Since all numbers $\overline{a b c d}, \overline{b c d a}, \overline{a b c d}, \overline{d a b c}$ are four-digit numbers, we have $1 \leqslant k \leqslant 9$ and $a, b, c, d \neq 0$.

We will sequentially analyze all cases. By subtracting the left and right sides of equation (I) from the left and right sides of the equality $10 \cdot \overline{a b c d}=\overline{a b c d 0}$, we get

$$
(10-k) \overline{a b c d}=9999 a
$$

Since $9999=3^{2} \cdot 11 \cdot 101$, both sides of equation (2) are divisible by the prime number 101, and due to the constraints $1 \leqslant 10-k \leqslant 9$, we get $\overline{a b c d}: 101 \Leftrightarrow \overline{a b}=\overline{c d}$. Dividing equation (2) by 101, we get

$$
(10-k) \overline{a b}=99 a
$$

Now, considering that both sides of the obtained equation are divisible by 11, and again due to the inequalities $1 \leqslant 10-k \leqslant 9$, we get $\overline{a b}: 11 \Leftrightarrow a=b$. Thus, the solutions to equation (I) are exactly all numbers of the form $\overline{\text { aaaa }}$ (for $k=1$).

Subtract equation (II) from the equality $100 \cdot \overline{a b c d}=\overline{a b c d 00}$. We get

$$
(100-k) \overline{a b c d}=\overline{a b} \cdot 9999
$$

Similarly to case (I), we have $\overline{a b c d}: 101 \Leftrightarrow \overline{a b}=\overline{c d}$. Dividing equation (3) by 101, we get

$$
(100-k) \overline{a b}=\overline{a b} \cdot 99
$$

Therefore, the solutions to equation (II) are exactly all numbers of the form $\overline{a b a b}$ (for $k=1$). Subtract equation (III) from the equality $1000 \cdot \overline{a b c d}=\overline{a b c d 000}$. We get

$$
(1000-k) \overline{a b c d}=\overline{a b c} \cdot 9999
$$

For $1 \leqslant k \leqslant 9$, we have $\text{GCD}(101, (1000-k))=1 \Rightarrow \overline{a b c d} \vdots 101 \Leftrightarrow \overline{a b}=\overline{c d}$. Dividing equation (4) by 101, we get

$$
(1000-k) \overline{a b}=\overline{a b a} \cdot 99
$$

Further, for $1 \leqslant k \leqslant 9$, we have $\text{GCD}(11, (1000-k))=1 \Rightarrow \overline{a b} \vdots 11 \Leftrightarrow a=b$. Therefore, the solutions to equation (III) are exactly all numbers of the form $\overline{\text { aaaa }}$ (for $k=1$).

Obviously, all solutions to equations (I) and (III) are contained among the solutions to equation (II).