Is it reducible?

For $\dfrac{n-17}{6n+11}$ not to be in lowest terms, it must be reducible, which implies that both the numerator and denominator must be divisible by a common factor, let's call it $m$ .

So we have, $m \vert (n-17)$ and $m \vert (6n+11$ ) $\;$ $\;$ $\;$ ( Note: $a \vert b\implies$ $a$ divides $b)$ .

Since $m \vert n-17$ and $m \vert 6n+11$ , $m\vert (6n+11)-6(n-17)$ $\implies m \vert 113$ But observe that $113$ is prime, so the only positive integers that divide $113$ are $1$ and $113$ itself . But $m$ cannot be $1$ and thus has to be $113$ . Plugging this back to the first equation, $113 \vert n-17$ . In order to achieve the minimum possible of $n$ , $n-17=113\implies n=\boxed{130}$