Solving System of Congruences Solve 
$x^{} \equiv 5 \pmod {7}$
$x^{} \equiv 5 \pmod {9}$
$x^{7} \equiv 5 \pmod {32}$
This is what I have done so far: 
$x=7k+5$
Substitute in x in the second congruence statement 
$7k+5 \equiv 5 \pmod {9}$
Solving this results in 
$k \equiv 0 \pmod {9}$
then
$k=9j$
$x=63j+5$
I'm lost from here on. I don't know how to deal with the $x^7$ congruence statement. 

By Euler's totient theorem, for all odd integer $x$:
$$x^{16} \equiv 1 \pmod{32}$$
Trivially, $x$ is odd, for if $x$ is even, then so would be $x^7$.
Then,
$$\begin{array}{rcll}
x^7 \equiv 5 & \pmod{32} \\
(x^7)^7 \equiv 13 & \pmod{32} \\
x^{49} \equiv 13 & \pmod{32} \\
x^{16\times3+1} \equiv 13 & \pmod{32} \\
(x^{16})^3 \cdot x \equiv 13 & \pmod{32} \\
(1)^3 \cdot x \equiv 13 & \pmod{32} \\
x \equiv 13 & \pmod{32} \\
\end{array}$$
You should be able to continue now.
