Order of $2^{36} \pmod{107}$ What is the order of $2^{36}\pmod{ 107}$? 
My current thought is $$2^{106} \equiv 1 \pmod{107}$$ according to Euler's Theorem. However, I don't know how to proceed from here or maybe my approach is wrong from the beginning?

You are correct that $(2^{36})^{106} \equiv 1\mod 107$.
So that means the order of $2^{36}$ divides $106$. (so is $1, 2, 53,$ or $106$)
But also not that $2^{106}\equiv 1 \mod 107$ so the order of $2$ also divides $106$.
If the order of $2^{36}$ were $1$ or $2$ the $2^{36}$ or $2^{72}$ would be equivalent to $1 \mod 107$ so that would mean the order of $2$ would divide both $106$ and $72$.  That would mean the order of $2$ would be $1$ or $2$.
It's easy to verify it is not.
So the order of $2^{36}$ must be $53$ or $106$.  As $(2^{36})^{53}=(2^{106})^{18} \equiv 1 \mod 106$ the order of $2^{36}$ is at most $53$ so it is $53$.
...
Another way of putting is if $|c^a| = k$ and $|c| = m$ then $c^{ak}\equiv 1\mod p$ ($p$ is prime) so $m|ak$ and $ak|p-1$ and $m|p-1$. 
That condition $m|ak$ and both $ak$ and $m$ dividing $p-1$ is very limiting.
$|2|| 36k$ and $36k|106$ means $18k|53$ means $k|53$ so $k|53$ so $|2|| 36k$ and $|2||106$.  
So $k = 53$ or $1$ and $|2| = 1,2,k,2k$.  As $|2|$ obviously does not equal $1$ and $2$, $k =53$ is the only option.
