TITLE: Sum of factors of a huge number.
QUESTION [10 upvotes]: I recently appeared in a math olympiad and it had this one question which had me stumped. This was a few weeks back and I have been looking for a way to find its answer ever since, but with no success. Searched the internet for the solution, but couldn't find any on it too! Anyway, here's how the question goes:
The value of $2^{96} - 3^{16}$ has two factors between 60 and 70. What is the sum of these two factors?
BTW, I should add that I did use wolframalpha to actually find the answer so I am more interested in knowing how to work it out manually than just knowing the answer. Any feedback would be appreciated.
Thanks!

REPLY [3 votes]: Modulo $61$:
$$2^6=64=3,2^{12}=9,2^{24}=81=20,2^{48}=400=34,\color{green}{2^{96}}=1156=\color{green}{58},$$
$$3^2=9=2^{12},\color{green}{3^{16}}=2^{96}=\color{green}{58}.$$
Similarly, modulo $67$ yields twice $25$.

Actually there is no need to perform the whole computation. Just observe
$$2^{12}\equiv3^2\mod61,\\2^{12}\equiv3^2\mod67.$$