Proof that all groups of order 616 are solvable.
I already know (counting elements) that at least one of the 2-, 7- or 11-sylow-subgroups has to be normal. Unfortunately i can not figure out which one. I think it might be possible to calculate each and every possibility but hope that someone can instead help with a clever way of solving this.
There are either $\;1\;$ or $\;56\;$ Sylow $\;11\,-$ subgroup. If the latter, then we already have $\;56\cdot10=560\;$ elements of order $\;11\;$. If the Sylow $\;7\,-$ subgroup isn't normal, then there are $\;8\;$ of them, which give $\;8\cdot6=48\;$ elements of order $\;7\;$, and thus there are left $\;616-560-48=8\;$ elements, which must thus a unique Sylow $\;2\,-$ subgroup.
Thus, there's always a normal Sylow subgroup, and taking the quotient by it we get a group of order $\;56,\,77\;$ or $\;88\;$, all of which are solvable. Of course, the normal Sylow subgroup is also solvable so we're done.