XI OBM

We show that $f(n)=91$ for $n\le 100$. For $n=100,99,\dots ,90$, we have $f(n)=f(f(n+11))=f(n+11-10)=f(n+1)$. But $f(101)=91$, so $f(n)=91$ for $n=100,99,\dots ,90$. Now for $n=89,88,\dots ,1,0,-1,\dots$ we use induction. We have $f(n)=f(f(n+11))=f(91)$ which, by induction, is 91. So the range of $f$ is all the integers greater than or equal to 91.