Group of order $630$ is solvable I tried using the Sylow theorems. $|G|=630=2\cdot 3^2 \cdot 5 \cdot 7$.
Denot the number of $p$-Sylow groups by $k_p$ then:
$k_2 \in \{1, 3, 5, 7, 9, 15, 21, 35, 45, 63, 315\}$
$k_3 \in \{1, 10, 70\}$
$k_5 \in \{1, 6, 21, 126\}$
$k_7 \in \{1, 15\}$
What are restrictions I might have to narrow the number of options I have? How do I continue from here?
 A: As I have explained in the comments, a trick using the Cayley embedding and  the Feit-Thompson theorem imply that every group whose order is not divisible by $4$ is solvable.
Here's a more elementary solution: we apply the same trick as in the comments: let $G$ be a group of order $2k$, where $k$ is odd, then consider the Cayley embedding $i:G \to S_{2k}$. The image of an element of order $2$ is a product of $k$ transpositions and thus an odd permutation. It follows that $i(G)$ is not contained in $A_{2k}$ and hence $i^{-1}(A_{2k})$ is a normal subgroup of index two (i.e. order $k$).
Thus for the question it suffices to show that a group of order $315=3^2 \cdot 5 \cdot 7$ is solvable. So let $G$ be of order $315$. If $n_3(G)=1$, then we are done, because then the $3$-Sylow is normal and hence we are reduced to showing that a group of order $35$ is solvable. But all groups with order $pq$ for $p,q$ prime are solvable, so we are done.
So suppose that $n_3(G)=7$. Then let $P$ be a $3$-Sylow subgroup of $G$, then we have $|N_G(P)|=45$. We know that either $P \cong C_9$ or $P \cong C_3 \times C_3$. In the first case $\mathrm{Aut}(P) \cong (\Bbb Z/9\Bbb Z)^\times$, which has order $\varphi(9)=6$. In the latter case, we have $\mathrm{Aut}(P) \cong \mathrm{GL}(2,3)$, which has order $48$. None of these is divisible by $5$, thus we get that there is a $5$-Sylow subgroup $Q$ of $G$ such that $Q \leq C_G(P)$ and hence $P \leq N_G(Q)$. From this it follows that $n_5(G)$ is not divisible by $3$, which implies that $n_5(G) = 1$. Thus there's a normal subgroup $Q$ of order $5$ and $G/Q$ has order $63=3^2 \cdot 7$. So it only remains to show that groups of order $63$ are solvable. I'll leave that as an exercise.
