If $G$ is a group of order $250,000 = 2^4 5^6$, show that $G$ is not simple. If $G$ is a group of order $250,000 = 2^4 5^6$, show that $G$ is not simple.
By the Sylow theorem, we have that the number of $2$-sylow subgroups of $G$ $n_2$ satisfy that
$$
n_2 \equiv1\mod2\mbox{ and } n_2\mid5^6
$$ 
Similarly for $n_5$ we have, 
$$
n_5 \equiv1\mod5\mbox{ and } n_5\mid2^4
$$
Hence,
$$
n_2 \in \{1,5,5^2,5^3,5^4,5^5,5^6\}, \mbox{ and } n_5 \in \{1,16\}
$$
But assuming that none of the $n_p'$s are one and using sylow theorems, I can't surpass the order of $G$ as the professor show us in class with one example. Now I am pretty sure I will need another approach but nothing comes to my mind. Any help would be appreciated.

As you state, $n_5$ is either $1$ or $16$. If it's the former, we are done. If it is $16$, then there is a homomorphism from $G\to S_{16}$ given by the fact that $G$ acts on the $5$-sylow subgroups by conjugation. But notice that the prime factorization of $|S_{16}|=16!$ contains exactly three copies of $5$ (coming from $5$, $10$ and $15$), but $|G|=2^45^6$ contains six copies of $5$. So this homomorphism cannot be injective so that it's kernel is a normal subgroup of $G$. So $G$ is not simple.
This expounds the well known aphorism that "Groups, as men, will be known by their actions".
