So I'm trying to find $9^{65}\pmod{77}$.
What I've done:
I used Fermat's little theorem and separated them into $9^{65}\pmod7$ and $9^{65}\pmod{11}$, where I got the remainders to be $4$ and $1$ respectively. Now, I have no idea what to do from this point. What is the best way to combine both to get the final answer?
I will show how to solve the congruences in a simple way that can be used for smaller values:
$$\begin{align} x&\equiv4 \pmod 7\tag1 \\x &\equiv 1 \pmod {11}\tag2 \end{align}$$
Now note that from $(1)$, $x=7a+4$, where $a$ is an integer. From $(2)$, $x \equiv 1 \pmod {11}$ so:
$$\begin{align} 7a+4 &\equiv 1 \pmod {11} \\7a &\equiv 8 \pmod {11} \\7a &\equiv 63 \pmod {11} \\a &\equiv 9 \pmod {11} \end{align}$$ So from the above congruence, $a=11b+9$ where $b$ is an integer. Therefore, $$x=7a+4 = 7(11b+9)+4 = 77b+67$$ Because $b$ is an integer, we can deduce that $$x \equiv 67 \pmod {77}$$
Use the extended Euclidean algorithm: $11x+7y=1$ , find $x = 2 , y = -3 $.
$9^{65}$ = $ (7y)+4(11x$) mod 77
Basic plug in
$9^{65}$ = 67
