<s>[INST] What is $\gcd(n-1,n+10)$, for $n$ a natural number? Currently I have,
let $d=\gcd(n-1,n+10)$
then, $d\mid n-1$ and  $d\mid n+10$
and hence $d$ must also divide $11$
not really sure what to do now, any help would be greatly appreciated.
 [/INST] As you have already observed, $d$ must divide $11$, thus $d$ is either $1$ or $11$. 
However, one thing to note that is both are possible-for example, for $n=2$, $d=1$, and for $n=12$, $d=11$. 
A similar idea would be using the Euclidean Algorithim, which yields that $$\gcd(n-1,n+10)=\gcd(n-1,n+10-n+1)=\gcd(n-1,11)$$ which is $1$ or $11$.
EDIT
An important observation would be noting when it is $1$, and when it is $11$. 
This matter can be resolved using our upper equation, which says that $\gcd(n-1,n+10)=\gcd(n-1,11)=11$. This implies that $11|n-1$, further showing that $\gcd(n-1,11)=11 \Rightarrow 11|n-1 \Rightarrow n \equiv 1 \pmod {11}$. 
Thus, $\gcd(n-1,11)=11$ implies that $n \equiv 1 \pmod {11}$. Further note that the converse is true-if $n \equiv 1 \pmod {11}$, then $n=11k+1$. 
This gives us $$\gcd(n-1,11)=\gcd(11k,11)=11$$  Thus we have $$\gcd(n-1,n+10)=11 \Leftrightarrow n \equiv 1 \pmod {11}$$
Since $\gcd(n-1,n+10)$ is either $1$ or $11$, this gives us that $$\gcd(n-1,n+10)=1 \Leftrightarrow n \not \equiv 1 \pmod {11}$$
This can be demonstrated in my upper example-if $n=2$, then $n \not \equiv 1 \pmod {11}$ thus implying that $\gcd(n-1,n+10)=1$. 
If $n=12$, then $n \equiv 1 \pmod {11}$, thus implying $\gcd(n-1,n+10)=11$. 
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