Let $G$ be a finite group. Let $\pi(G)=\{2,3,5\}$ be the set of prime divisors of its order. If 6 divide the number of Sylow 5-subgroups of G and 10 divide the number of Sylow $3$-subgroups of $G$, then whether the group $G$ group with those properties is unsolvable?
In particular if the number of Sylow $5$-subgroups of $G$ is 6 or the number of Sylow $3$-subgroups of $G$ is 10, then by the Hall's theorem $G$ is unsolvable group. For example if the number of Sylow $5$-subgroups of $G$ is 6 and $G$ is solvable, then $2\equiv 1$ (mod $5$), a contradiction.
Hall's theorem: Let $G$ be a finite soluble group and $|G|=m.n$, where $m=p_{1}^{\alpha _{1}}...p_{r}^{\alpha _{r}}$, $(m,n)=1$. Let $\pi =\{p_{1},...,p_{r}\}$ and $ h_{m}$ be the number of $\pi -$Hall subgroups of $G$. Then $ h_{m}=q_{1}^{\beta _{1}}...q_{s}^{\beta _{s}}$, satisfies the following conditions for all $i\in \{1,2,...,s\}$:
1) $q_{i}^{\beta_{i}} \equiv 1$ (mod $p_{j}$), for some $p_{j}$.
2) The order of some chief factor of $G$ is divisible by $ q_{i}^{\beta_{i}}$.
Thank you so much.