If $j$, $k$, and $n$ are consecutive integers such that $0
(A) $0$
(B) $1$
(C) $2$
(D) $3$
(E) $4$
I'm thinking the answer is (B) because the only factors of $9$ is $1,\, 9$, and $3$ ($\large1\times9\text{ and }3\times3$) so we can take out (A), (C), and (E). All that's left is (B) and (D). But since $k$ is less than $n$, they both can't be $3$ for $jn=9$. And that leaves (B) as the answer?
Am I right in doing it this way?