jmc

Let $d=\gcd (n+5,n+11)$, so $d$ divides both $n+5$ and $n+11$. Then $d$ divides $(n+11)-(n+5)=6$. Therefore, $d$ can only be 1, 2, 3, or 6.

If $n=2$, then $\gcd (n+5,n+11)=\gcd (7,13)=1$.

If $n=3$, then $\gcd (n+5,n+11)=\gcd (8,14)=2$.

If $n=4$, then $\gcd (n+5,n+11)=\gcd (9,15)=3$.

If $n=1$, then $\gcd (n+5,n+11)=\gcd (6,12)=6$.

Hence, all the values 1, 2, 3, and 6 are attainable, for a total of $\boxed{4}$ possible values.