Number theory: divisors of $|2016^m - 36^n|$. Let $N$ be a natural number and $p$ be the number of its divisors. If we know that $N =|2016^m - 36^n| $, what is the minimum number $p$ if $m$ and $n$  are natural numbers ($m,n > 0$)?
 A: We know the answer is at most $30$ since $2016-36^2=720$ has $30$ divisors. I will prove it is optimal.
It is trivial to see that for any integer $k$, if $k$ divides $N$ then $N$ has same or more divisors than $k$.
Firstly, $5$ always divides $N$. And if $\min(m,n)\ge 2$, then $36^2=2^43^4|N$. Therefore, $N$ has at least $50$ divisors (from $2^43^45$). We can exclude this case and suppose that $m=1$ or $n=1$.
i) $m=1$: For $n<3$, we can check manually that there are no $N$ with less than $30$ divisors. For $n \ge 3$, we can see that $1440=2^53^25|N$ so $N$ has at least $36$ divisors.
ii) $n=1$: $180=2^23^25|N$, so $N$ has at least $18$ divisors. If there is another prime factor of $N$, we are done. Let's suppose that there are no other prime factor of $N$, and also $m>1$ (we can check $m=1$ case manually).
We can observe that since $2^3|2016^m$ and $2^3$ does not divide $36$, $2^3$ does not divide $N$. Similarly, $3^3$ does not divide $N$. For $N$ to have less than or equal to $30$ divisors, $N \le 2^23^25^3=4500$ and some calculation shows that this is impossible.
Therefore, we can conclude that $p=30$.
