Using the Chinese Remainder Theorem to solve the following linear congruence: $17x \equiv 9 \pmod{276}$ The book I am following (Elementary Number Theory by David Burton) uses the Chinese Remainder Theorem to solve $17x \equiv 9 \pmod{276}$ by breaking it up into a system of three linear congruences,
$$17x \equiv 9 \pmod{3}$$
$$17x \equiv 9 \pmod{4}$$
$$17x \equiv 9 \pmod{23}$$
I realize that the latter system is guaranteed to have a unique solution modulo $3 \times4\times23 = 276$. What I fail to understand is how and why the solution of this system of congruences the same as the solution of the initial congruence, $17x \equiv 9 \pmod{276}$. How can I solve this linear congruence using the Chinese Remainder Theorem?
 A: First Bit - why the solution of the system is a valid answer
If we have a number $y$ such that $y\equiv9\pmod{3}$ and $y\equiv9\pmod{4}$ and $y\equiv9\pmod{23}$ then we know that $y-9\equiv0\pmod{3}$ and $y-9\equiv0\pmod{4}$ and $y-9\equiv0\pmod{23}$. In other words $y-9$ divides by $3$, $4$ and $23$. As these are all coprime then $y-9$ must divide by their product which is $276$.
Second Bit - actually doing the math
$17x\equiv9\pmod{3}$
$17x\equiv0\pmod{3}$
So $x=3y,y\in\mathbb{Z}$
So the second equation becomes
$17x\equiv9\pmod{4}$
$51y\equiv1\pmod{4}$
$3y\equiv1\pmod{4}$
$y\equiv3\pmod{4}$
So $y=4z+3,z\in\mathbb{Z}$
Putting this into the third equation:
$17x\equiv9\pmod{23}$
$51y\equiv9\pmod{23}$
$5y\equiv9\pmod{23}$
$5(4z+3)\equiv9\pmod{23}$
$20z+15\equiv9\pmod{23}$
$20z+6\equiv0\pmod{23}$
This leads easily to $z=2$ which then gives $y=11$ and hence $x=33$.
Checking: $17\times33=561=2\times276+9$
A: From the first congruence you have that $ x \equiv 0  \pmod 3$. The second one gives us $ x \equiv 1 \pmod 4$ and the third one, by multiplying with $-4 \pmod{23}$  gives us that $ x \equiv 10 \pmod{23}$ .
Now clearly the unique solution of the first two congruences is $ x \equiv 9  \pmod{12}$
So it remains to solve the system $$ x \equiv 9 \pmod{12} $$ $$ x \equiv 10 \pmod{23} $$
Let $ M_{1}=23 $ and $ M_{2}=12 $. We solve the congruences $$ M_{1}y_{1} \equiv 1 \pmod{12} $$
$$ M_{2}y_{2} \equiv 1 \pmod{23} $$
Take $ y_{1} $ and $ y_{2} $ to be the smallest positive integers satisfying this congruences.In our case we obtain $ y_{1}=11 $ and $ y_{2}=2 $. Then the solution of the original system of congruences is $$ x \equiv 9M_{1}y_{1}+10M_{2}y_{2} \pmod{276}$$
After computation, we obtain $$ x \equiv 33\pmod{276} $$
A: This is because the canonical map $\;\mathbf Z/276\mathbf Z\to\mathbf Z/3\mathbf Z\times\mathbf Z/4\mathbf Z\times\mathbf Z/23\mathbf Z\;$ is an isomorphism.
To solve the congruence you solve each congruence separately. Then you find the solution progressively. If you find, say, $x\equiv a\mod 4$ and $x\equiv b\mod 23$, you find the solution modulo $4\times 23$ from a *Bézout's relation between $4$ and $23$.  Here you obviously have
$$6\times 4-23=1,$$
and in the general case, you have to use the extended Euclidean algorithm. From this relation, you obtain the solution 
$$x\equiv  b\cdot6\times4-a\cdot2\mod 4\times23.$$
Then do the same operations with $3$ and $92$.
