10. Let $A$, $B$, $C$, $D$ be four different natural numbers less than 10, and satisfy:
$$
\begin{array}{rrrr}
A B B C B \\
+B C A D A \\
\hline D & B & D & D
\end{array}
$$

Then $D$ has ( ) different values.
(A) 2
(B) 4
(C) 7
(D) 8
(E) 9
10. C.

From the problem, we know that $C=0, A+B=D$.
Since $A, B, D$ are different digits, and $A, B$ are the leading digits, therefore, $A, B$ are not 0.
Thus, $3=1+2 \leqslant A+B=D \leqslant 9$.
Therefore, $D=3,4,5,6,7,8,9$. There are seven different values in total.