Show that every group of order $2 · 3 · 5 · 67$ has a normal subgroup of order $5$.

Question

Show that if $G$ is a group of order $2010 = 2 · 3 · 5 · 67$, then $G$ has a normal subgroup of order $5$.

I tried use Sylow's theorem to show that $k_5=1$. $k_5$ is equal to $1$, $6$, or $201$.