What is the remainder when $24^{1202}$ is divided by $1446$?

Question

I tried remainder theorem but that does not simplify it.

I tried factorizing $1446$ as $2\cdot3\cdot241$ and got remainders when numerator is divided by $2,3$ and $241$ individually but then I did not know what to do next? Stuck!

Answer 1

As $(24,1446)=6$

let use find $\displaystyle24^{1201}\pmod{241}$

Now, $\displaystyle\phi(241)=240$ and $1201\equiv1\pmod{240}\implies24^{1201}\equiv24^1\pmod{241}$

Now use $a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod {m\cdot c} $

$\displaystyle\implies24^{1201}\cdot24\equiv24^1\cdot24\pmod{241\cdot24}\equiv576\pmod{241\cdot6} $