Find the solution of the congruence modulo 4199 I am really bummed out on this problem, can't find any helpful links. But the problem asks:
1.) Find the solution of the congruences modulo 4199:
$x ≡ 11 \ (mod \ 13), \ x ≡ 5 \ (mod \ 17), \ x ≡ 17 \ (mod \ 19)$
I really need a good explanations on solving this problem, if anyone can help out, that would be appreciated. 
 A: For two congruences involving coprime moduli:
Start from a Bézout's relation between the moduli: $\;um+vn=1$.
$$\begin{cases}
x\equiv a\pmod m\\x\equiv b\pmod n
\end{cases}\iff x\equiv bum+avn\pmod{ab}.$$
For more than two congruences modulo pairwise coprime moduli: solve the first two congruences,then solve the system made up of  the resulting congruence and the third  congruence, and so on
A: Let's write the solution as:
$$x = 11 A + 5 B + 17 C\tag{1}$$
where we choose $A$ such that it equals $1$ when reduced modulo $11$ but it is zero when evaluated modulo $17$ and $19$ and we choose $B$ and $C$ analogously such that $x$ manifestly satisfies the 3 congruence relations. So, all we need to know is how to find a number that's equal to $1$ modulo some given number $y_1$ while it's zero modulo some other numbers $y_2, y_3$ etc. that are relatively prime. The product of the numbers $y_2, y_3,\ldots$ will clearly be zero modulo these numbers, and if you multiply that by the inverse of that number modulo $y_1$ it will obviously remain zero modulo  $y_2, y_3,\ldots$, but it will now be equal to $1$ modulo $y_1$.
We can thus write down the coefficients in Eq. (1) as:
$$
\begin{split}
A &= 17\times 19 \times \left(\left(17\times 19\right)^{-1}\bmod 13\right) =17\times 19\times 6 \\
B &= 13\times 19 \times \left(\left(13\times 19\right)^{-1}\bmod 17\right) =  13\times 19 \times 2\\
C &= 13\times 17 \times \left(\left(13\times 17\right)^{-1}\bmod 19\right) =  13\times 17 \times 8
\end{split}
$$
This method is known as the Chinese remainder theorem, but i.m.o., it's best to use the logic behind the theorem to solve the problem rather than to just look it up and use the formula when you are not yet familiar with the details.
A: We can see immediately that for the $\bmod 13$ and $\bmod 19$ cases we have 
$\left .
\begin{array}r
x\equiv 11\equiv \color{blue}{-2} \bmod 13 \\
x\equiv 17\equiv \color{blue}{-2} \bmod 19 \\
\end{array}
\right \}
\implies x\equiv \color{blue}{-2}\equiv 245 \bmod 247$
Then simultaneously solving $x\equiv 5 \bmod 17$ needs $17^{-1} \bmod 247$, which just for variety I'll solve in equations:
$\begin{align}
 &&17a &\equiv 1 \bmod 247 \\
\small(\times 15) &&255a\equiv 8a  &\equiv 15 \bmod 247 \\
\small(\times 31) &&248a\equiv a  &\equiv 365 \equiv -29 \bmod 247 \\
\end{align}$
giving 
$\begin{align}
17k + 5 &\equiv -2\bmod 247 \\
17k  &\equiv -7 \bmod 247 \\
k  &\equiv 7\cdot 29 \equiv 203 \bmod 247 \\
\end{align}$
Then $x\equiv 17\cdot 203+5 \equiv  3456 \bmod 4199$
