Let's find different digits for the letters $A, B, C, D$ such that the following division holds in the decimal system:

$$
A B C: B B B B=0, B C D B \quad B C D B \quad \ldots
$$

(i.e., the quotient should be a repeating decimal).
Any ordinary fraction can easily be converted into a decimal fraction by anyone. Conversely, a non-repeating decimal fraction that does not consist entirely of 9s can always be represented as the quotient of two integers, such that the numerator of the fraction contains the digits of the segment, and the denominator is \(10^k - 1\), where \(k\) is the number of digits in the segment, e.g., \(0.873873 \ldots = \frac{873}{999}\). Indeed, instead of the usual four steps of dividing \(BCDB\) by 9999, the dividend can be written as \(BCDB \cdot 10^4\), and this can be expressed as \(BCDB \cdot 9999 + BCDB\), so both the quotient and the remainder are \(BCDB\), and the process repeats:

\[
BCDB: 9999 = 0, BCDB \quad BCDB \ldots
\]

(since \(B \neq C\) implies \(BCDB1000 \cdot B^2\), and from the inequality

\[
9000 > B^2 \cdot 1000
\]

it follows that \(B\) can be 1 or 2.

First, let's choose \(B\) to be 1. From (1),

\[
9 \cdot (A1C) = 1 \cdot (1CD1)
\]

The multiplication \(9 \cdot C = \ldots 1\) holds only if \(C = 9\), and then

\[
\frac{19D1}{A19} = 9 \text{ implies }
\]

\(A = 2\) and \(D = 7\) are obtained, and indeed \(219: 1111 = 0.19711971 \ldots\)

In the case of \(B = 2\), from (1),

\[
9(A2C) = 2(2CD2) \text{ and } 9C = \ldots 4 \text{ implies only } C = 6. \text{ From this, }
\]

\[
\begin{gathered}
\mid 9(100A + 20 + 6) = 2(26D2) = 5204 + 20D \\
900A + 234 = 5204 + 20D \\
90A - 2D = 497
\end{gathered}
\]

On the left side, we expect integers for \(A\) and \(D\), but the difference cannot be odd, so there is no solution. Therefore, the found solution is unique.