Find remainder when $20^{13}$ is divided by $4940$

Question

Find remainder when $20^{13}$ is divided by $4940$

I have solved this, but am hoping for an elegant solution.

My solution:

$r(20^{13}||4940) = 20 \times r(20^{12}||247)$ where $r(a||b)$ is the remainder when $a$ is divided by $b$.

$20^3 \equiv 96 \mod 247 $

$20^6 \equiv 77 \mod 247 $

$20^{12} \equiv 1 \mod 247 $

So my answer is $20$.

But there must be a better way (This seems kind of brute force).

Any hints are welcome.

Answer 1

As $247=13\cdot19,$

Using Fermat's Little Theorem, $20^{13-1}\equiv1\pmod{13}$

and $20\equiv1\pmod{19}\implies20^{12}\equiv1^{12}\equiv1$

$\implies20^{12}\equiv1\pmod{\text{lcm}(13,19)}$