In the equation below, $A$ and $B$ are consecutive positive integers, and $A$, $B$, and $A+B$ represent number bases: \[132_A+43_B=69_{A+B}.\]
What is $A+B$?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17$
Change the equation to base 10: \[A^2 + 3A +2 + 4B +3= 6A + 6B + 9\] \[A^2 - 3A - 2B - 4=0\]
Either $B = A + 1$ or $B = A - 1$, so either $A^2 - 5A - 6, B = A + 1$ or $A^2 - 5A - 2, B = A - 1$. The second case has no integer roots, and the first can be re-expressed as $(A-6)(A+1) = 0, B = A + 1$. Since A must be positive, $A = 6, B = 7$ and $A+B = \boxed{\textbf{(C)}\ 13}$