In the addition shown below $A$, $B$, $C$, and $D$ are distinct digits. How many different values are possible for $D$?
\[\begin{array}[t]{r}     ABBCB \\ + \ BCADA \\ \hline     DBDDD \end{array}\]

$\textbf {(A) } 2 \qquad \textbf {(B) } 4 \qquad \textbf {(C) } 7 \qquad \textbf {(D) } 8 \qquad \textbf {(E) } 9$
Note from the addition of the last digits that $A+B=D\text{ or }A+B=D+10$. 
From the addition of the frontmost digits, $A+B$ cannot have a carry, since the answer is still a five-digit number. Also $A + B$ cant have a carry since then for the second column, $C + 1 + D$ cant equal $D$. 
Therefore $A+B=D$. 
Using the second or fourth column, this then implies that $C=0$, so that $B+C=B$ and $C+D=D$. 
Note that all of the remaining equalities are now satisfied: $A+B=D, B+C=B,$ and $B+A=D$. Since the digits must be distinct, the smallest possible value of $D$ is $1+2=3$, and the largest possible value is $9$. Thus we have that $3\le D\le9$, so the number of possible values is $\boxed{\textbf{(C) }7}$