The Chinese remainder theorem I'm trying to do some questions on the Chinese remainder theorem, I've being reading the Wikipedia explanation but I still don't get it. Can someone explain it to me, please?
Here is the question I'm working with. If you have a simpler example, that would be great.

Find an integer $x$ such that $x\equiv2\bmod5$, $x\equiv1\bmod13$ and $x\equiv5\bmod17$.


$$\begin{align}
   x &\equiv 2 \pmod5 \\
   x &\equiv 1 \pmod{13} \\
   x &\equiv 5 \pmod{17}
\end{align}$$

This is not the fastest way; but it is the most straightforward way. This method only works when the moduli are pairwise prime:
$$\gcd(5,13) = \gcd(5,17) = \gcd(13,17) = 1$$
We start with $ x \equiv 2 \pmod5$. Then for some integer, $u$, we must have
$$x = 5u+2$$ Then $x \equiv 1 \pmod{13}$ becomes
\begin{align}
   x &\equiv 1 \pmod{13} \\
   5u+2 &\equiv 1 \pmod{13} \\
   5u &\equiv -1 \equiv12 \equiv 25\pmod{13} 
      &\text{(Add $13$ until you get a multiple of $5$.)} \\
   u &\equiv 5 \pmod{13} \\
\hline
   u &= 13v + 5
\end{align}
So now we have $$x = 5u+2 = 5(13v+5)+2 = 65v + 27$$
Next we take $x \equiv 5 \pmod{17}$. (The first few multiples of $17$ are
$17, 34, 51, 68, 85$).
\begin{align}
   x &\equiv 5 \pmod{17} \\
   65v + 27 &\equiv 5 \pmod{17} \\
   -3v&\equiv -5 \pmod{17} \\
   3v &\equiv 5 \equiv 22 \equiv 39 \pmod{17} \\
   v &\equiv 13 \pmod{17} \\
\hline
   v &= 13 + 17w
\end{align}
So $x = 65v + 27 = 65(17w + 13)+27 = 1105w + 872$
Which we can write as
$$x \equiv 872 \pmod{1105}$$
