Solve $17x \equiv 3 \pmod {2\cdot 3 \cdot 5 \cdot 7}$ by solving the system $$17 x \equiv 3 \pmod 2 \qquad \qquad 17x \equiv 3 \pmod 3$$ $$17x \equiv 3 \pmod 5 \qquad \qquad 17x \equiv 3 \pmod 7$$
Edited:
I solved it using a method introduced in my class:
$N_1=5 \cdot 3 \cdot 7 = 105 \implies 105x_1 \equiv 1 \pmod 2 \implies x_1 \equiv 1 \pmod 2 \implies x_1 \equiv 1 \pmod 2$
$N_2=2 \cdot 5 \cdot 7 = 70 \implies 70x_2 \equiv 1 \pmod 3 \implies x_2 \equiv 1 \pmod 3 \implies x_2 \equiv 1 \pmod 3$
$N_3=2 \cdot 3 \cdot 7 = 42 \implies 42x_3 \equiv 1 \pmod 5 \implies 2x_3 \equiv 1 \pmod 5 \implies x_3 \equiv 3 \pmod 5$
$N_4=2 \cdot 3 \cdot 5 = 30 \implies 30x_4 \equiv 1 \pmod 7 \implies 2x_4 \equiv 1 \pmod 7 \implies x_4 \equiv 4 \pmod 7$
And I was given that the solution is given by the below formula \begin{align*} \bar x &= a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 + a_4 N_4 x_4 \\ &= 3 \cdot 105 \cdot 1 + 3 \cdot 70 \cdot 1 +3 \cdot 42 \cdot 3+ 3 \cdot 30 \cdot 4 \\ &= 1263 \\ &\equiv 3 \pmod{2 \cdot 3 \cdot 5 \cdot 7} \end{align*} But this is a wrong answer, so there is something wrong in my work but I can't find it. :( I checked an online calculator for the Chinese remainder theorem, and it returned me a different value. I think that value is $x\equiv 193 \pmod{2 \cdot 3 \cdot 5 \cdot 7}$.
OK, perhaps it is not fair to write an answer like this (but it is true): Certainly $N_3=2\cdot 3\cdot 7=70$ is wrong. And $N_2=2\cdot 5\cdot 7=42$, too, for that matter.
Edit: I get $x=99$. Then $17\cdot 99=1683\equiv 3\bmod 210$. All solutions are given by $x=99+210k$ with $k\in \mathbb{Z}$.