$45=5\times9$  $\overline{392AB}$ should be ended with $0$ or $5.$ But $0$ cannot be the first digit. Thus, $B$ can only be $5$ and $A$ should be $8.$ 
$\overline{392AB}$ is a multiple of $45$, and $\overline{B34}$ is a three-digit even number. What is the sum of $A$ and $B$? 
Options:
a: $$13$$ 
b: $$4$$ 
c: $$5$$ 
d: $$7$$ 
e: $4$ or $13$ 