Using the Chinese Remainder Theorem, $17x \equiv 9 \pmod{276}$ I want to uses the Chinese Remainder Theorem to solve $17x \equiv 9 \pmod{276}$ by breaking it up into a system of three linear congruences,
$$17x \equiv 9 \pmod{3}$$
$$17x \equiv 9 \pmod{4}$$
$$17x \equiv 9 \pmod{23}$$
For that I reduced it to
$$x \equiv 0 \pmod{3}$$
$$x \equiv 1 \pmod{4}$$
$$17x \equiv 9 \pmod{23}$$
So for converting this In terms of chinese reminder Theorem , I calculate The solution Of last linear Congurence as
$$x \equiv 13 \pmod{23}$$
So Our System Of Linear Congurence is now :
$$x \equiv 0 \pmod{3}$$
$$x \equiv 1 \pmod{4}$$
$$x \equiv 13 \pmod{23}$$
And now I apply the Chinese Remainder Theorem on it such that
$$92b_1 \equiv 1 \pmod{3}$$
$$69b_2 \equiv 1 \pmod{4}$$
$$12b_3 \equiv 1 \pmod{23}$$
So $b_1$ = 2 , $b_2$ = 1 , $b_3$ = 2
So simultaneous solution be
$$92\cdot2\cdot0 + 69\cdot1\cdot1 + 13\cdot2\cdot5 = 199$$
But it's wrong (@_@)༎ຶ‿༎ຶ . Can please Please Someone can Correct me.

$17x\equiv 9\bmod 276$
$276=16\times 17 +4\Rightarrow 17x=9+(16(17)+4)k$
$\Rightarrow 17(x-16k)=9+4k$
For $k=2$ we have:
$17(x-16\times2)=9+4\times 2=17$
$\Rightarrow x-32=1\rightarrow x=33$
$k=53\rightarrow x-53\times 16=13\times 17\rightarrow x=861$
K makes an arithmetic progression with common difference $d=51$:
$k= 2, 53, 104, 155\cdot\cdot\cdot$
