Determine the remainder of $2^{666}\pmod {2016}$ with the Chinese Remainder Theorem.
I started with 2016 = $2^{5}$.$3^{2}$.$7$
I made: $$x \equiv 2^{666} \pmod{2^{5}}$$ $$x \equiv 2^{666} \pmod9$$ $$x \equiv 2^{666} \pmod7$$
Because 2 and 9 are coprime and 2 and 7, I used Euler's totient function to simplify it.
$\phi (9)$ = 6 and $\phi (7)$ = 6.
666 = 111.6 + 0
So:
$$x \equiv 2^{666} \pmod{2^{5}}$$ $$x \equiv 1 \pmod9$$ $$x \equiv 1 \pmod7$$
But because 2 and $2^{5}$ are not coprime how could I simplify it and find the remainder with the Chinese Remainder Theorem?