4 digits

$\begin{array}{ c c c c c c c c c } & & A & B& C& D\\ & & D & A& B& C\\ & & C &D & A & B\\ + & & {\color{#D61F06}0} & C & D & A \\ \hline & C & C & C & C & B \end{array}$

Observe the cryptarithm carefully that the first four terms in the sum list some of the possible arrangements of $A, B, C$ and $D$ , where the thousands digit of $\overline{CDA}$ is 'invisibly' $B$ . Noting that $B = 0$ , we have $A + C + D \equiv 0 \bmod 10$ Since $A + C + D \geq 10$ , and $A + C + D$ exists in three remaining decimal places, there are $4$ leftmost carry digits of the same value $C$ , which is nonzero. So either $C = 1$ or $C = 2$ .

• If $C = 2$ , then $D + A \equiv 8 \bmod 10$ . Since the carry digit is $2$ , then $D = A = 9$ , which is not possible.
• If $C = 1$ , then $A + D \equiv 9 \bmod 10$ , so $A + D = 9$ . From here, we have $8$ distinct integers from $2$ to $9$ . Since $1$ and $0$ are already used, this eliminates the candidates $8$ and $9$ , which then reduces to $6$ integers from $2$ to $7$ . In this case, there are $7 - 2 + 1 = 6$ possible ways to express two integers as the sum of $9$ .

Thus, since the sum depends on the variables $A$ and $B$ , there are $1 \cdot 1 \cdot 6 = \boxed{6}$ possible choices to substitute $\overline{ABCD}$ .

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Note: I will check to see if there are other accurate/complete approaches for this problem. I believe my solution seems to be less rigorous because there can be other choices of $B$ that can also work.