$\begin{array}{ c c c c c c c c c } & & A & B& C& D\\ & & D & A& B& C\\ & & C &D & A & B\\ + & & & C & D & A \\ \hline & C & C & C & C & B \end{array}$

If $A, B, C, D$ are four distinct digits, how many possible values does $\overline{ABCD}$ have?

9
10
4
5
8
7
3
6

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

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 $C = 2$ .

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

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.