Let $G$ be a group of order $1331$. Prove that $G$ has at least $11$ elements of order $11$.
So by First Sylow's theorem, there exists a Sylow $11$-subgroup of G. By Third Sylow's theorem, the number of such subgroups is $11k+1$ and $11k+1|1331$, thus, this is only possible for $k=0$. This means that the Sylow $11$-subgroup is unique, and therefore there exist at least $10$ elements of order $11$ in $G$.
So it appears I'm missing one element to complete the proof. Have I done something wrong?