System of 3 Congruences Trying to solve this system of congruences:
$$n \equiv 4 \quad \text{(mod 19)}$$
$$n \equiv 3 \quad \text{(mod 10)}$$
$$n \equiv 6 \quad \text{(mod 11)}$$
how do I solve this?
 A: $$
\begin{eqnarray}
10\cdot11&\equiv&0&\mod10\;,\\
10\cdot11&\equiv&0&\mod11\;,\\
10\cdot11&\equiv&15&\mod19\;.
\end{eqnarray}
$$
Now you just need to solve $15x\equiv4\mod19$, and then $10\cdot11\cdot x$ solves the first congruence without disturbing the other two. Then you can do the same thing for the other two.
A: It's a simple application of Easy CRT (below). With practice, this can be calculated mentally.
$$\!\!\!\!\!\!\!\!\!\rm \begin{eqnarray}\rm n&\equiv&\rm 6\ (mod\ 11) \\
\rm n&\equiv&\rm 4\ (mod\ 19)\end{eqnarray}\!\!\! \iff\!\!  n \equiv 4 + 19\: \bigg[\!\!\frac{6\!-\!4}{19} mod\ 11\bigg]\! \equiv 61\  (mod\ 209)\ \ by\  \frac{2}{19}\equiv \frac{-9}{-3}\equiv 3 \ (mod\ 11)$$ 

$$\rm\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \begin{eqnarray}\rm n&\equiv&\rm\ 3\ \ (mod\ 10) \\
\rm n&\equiv&\rm 61\: (mod\ 209)\end{eqnarray}\!\!\! \iff\!\!  n \equiv 61 + 209 \bigg[\!\!\frac{3\!-\!61}{209} mod\ 10\bigg]\! \equiv -357\  (mod\ 2090)\ \ by\  \frac{2}{-1}\equiv -2\: (mod\ 10)$$
THEOREM (Easy CRT) $\rm\ \ $ If $\rm\ m,\:n\:$ are coprime integers then $\rm\ n^{-1}\ $ exists $\rm\ (mod\ m)\ \ $ and
$\rm\displaystyle\quad \begin{eqnarray}\rm x&\equiv&\rm\ a\ (mod\ m) \\
\rm x&\equiv&\rm\ b\ (mod\ n)\end{eqnarray}\!\!\! \iff\!  x\: \equiv\: b + n\ \bigg[\frac{a\!-\!b}{n}\ mod\ m\:\bigg]\ \ (mod\ m\:\!n)$
Proof $\rm\ (\Leftarrow)\quad mod\:\ n\!:\:\ x\:\equiv\: b + n\ (\cdots)\equiv b\:,\ $ and $\rm\ mod\ m\!:\ x\equiv b + (a-b)\ n/n\: \equiv\: a\:.$
$\rm\ (\Rightarrow)\ \ $ The solution is unique $\rm\ (mod\ m\:\!n)\ $ since if $\rm\ x',x\ $ are solutions then $\rm\ x'\equiv x\ $ mod $\rm\:m,n\:$ therefore $\rm\ m,n\ |\ x'-x\ \Rightarrow\ m\:\!n\ |\ x'-x\ \ $ since $\rm\ \:m,n\:$ coprime $\rm\:\Rightarrow\ lcm(m,n) = m\:\!n\ \ \ $ QED
