Remainder when ${45^{17}}^{17}$ is divided by 204 Find the remainder when ${45^{17^{17}}}$ is divided by 204
I tried using congruence modulo. But I am not able to express it in the form of $a\equiv b\pmod{204}$.
$204=2^2\cdot 3\cdot 17$
 A: Hint: By the Euler-Fermat theorem we know $45^{\varphi(4)} = 45^{2} \equiv 1 \mod 4$. 
By the Euler-Fermat theorem we also know $45^{\varphi(17)} = 45^{16} \equiv 1 \mod 17$ 
It is clear that $45 \equiv 0 \mod 3$ .
The Chinese Remainder Theorem gives us now $45^{16} \equiv 69 \mod 204$. Hence $45^{17} \equiv 69 \cdot 45 = 3105 \equiv 45 \mod 204$.
A: Hint:
$45\equiv1\pmod4\implies45^n\equiv1$
$45\equiv0\pmod3\implies45^m\equiv0$
$45\equiv11\pmod{17}$ 
and $17\equiv1\pmod{16}$ and $17^n\equiv1$
$\implies11^{17^{17}}\equiv11\pmod{17}$
Now apply CRT
A: HINT:
$45^{17}\equiv45\pmod{204}$
A: As $(45,204)=3$
let us start with $45^{17^{17}-1}\pmod{\dfrac{204}3}$
Using Carmichael function, $\lambda(204)=16$ and $17\equiv1\pmod{16}\implies17^{17}\equiv1$
$\implies45^{17^{17}-1}\equiv1\pmod{68}$
More generally, $a^{17^n-1}\equiv1\pmod{68}$  if $(a,68)=1$ and $n$ is a positive integer
Now, $45^{17^{17}-1}\cdot45\equiv1\cdot45\pmod{68\cdot45}$
Now using the fact: $204\mid68\cdot45,$
$$45^{17^{17}}\equiv45\pmod{204}$$
