If $a$ and $b$ are positive integers such that $\frac{1}{a}+\frac{1}{2 a}+\frac{1}{3 a}=\frac{1}{b^{2}-2 b}$, then the smallest possible value of $a+b$ is
(A) 8
(B) 6
(C) 96
(D) 10
(E) 50
Since $\frac{1}{a}+\frac{1}{2 a}+\frac{1}{3 a}=\frac{1}{b^{2}-2 b}$, then $\frac{6}{6 a}+\frac{3}{6 a}+\frac{2}{6 a}=\frac{1}{b^{2}-2 b}$ or $\frac{11}{6 a}=\frac{1}{b^{2}-2 b}$.

Cross-multiplying, we obtain $11\left(b^{2}-2 b\right)=6 a$.

Since 11 is a divisor of the left side, then 11 must be a divisor of the right side, that is, $a$ must be divisible by 11 .

Thus, let $a=11 A$, with $A$ a positive integer.

So we get $11\left(b^{2}-2 b\right)=6(11 A)$ or $b^{2}-2 b=6 A$.

Since 6 is a divisor of the right side, then 6 must be a divisor of the left side.

What is the smallest positive integer $b$ for which 6 is a divisor of $b^{2}-2 b$ ? We can quickly check that if $b$ equals $1,2,3,4$, or 5 , then $b^{2}-2 b$ is not divisible by 6 , but if $b=6$, then $b^{2}-2 b$ is divisible by 6 .

Therefore, for the smallest values of $a$ and $b$, we must have $b=6$, so $6 A=6^{2}-2(6)=24$, whence $A=4$ and so $a=11 A=44$.

Thus, the smallest possible value of $a+b$ is $44+6=50$.

ANSWER: (E)