Given a group $G$ with $|G|=351$, prove that $G$ is solvable.
my attempt:
From the first Sylow Theorem, we know that if we set $P_3,P_{13}$ the sylow groups of $3,13$ then $n_{13}\in\{1,27\},n_3\in\{1,13\}$ and I know that if $n_{13}=27$ then $n_3=1$ so we can set the following normal series
$G\rhd P_3\rhd{e}$ and $G/P_3$ is of order 13 and then cyclic and abelian but I wasn't sure about $P_3/\{e\}$.
No need to prove it by definition, just use the following theorem: if $G$ contains a normal subgroup $N$ such that both $N$ and $G/N$ are solvable then $G$ is solvable as well. One of the corollaries of this theorem is that $p$-groups are solvable.
So in your case, you know that either $n_3=1$ or $n_{13}=1$. This means $G$ contains a normal Sylow-subgroup $N$. We have either $|N|=13, |G/N|=27$ or vice versa. In either case, both $N$ and $G/N$ are solvable.