Solve simultaneous system of congruence
$$\text{$x\equiv 2\pmod{910} \qquad$ and $\qquad x\equiv 93\pmod{1001}$}$$
I tried Chinese remainder theorem on this, but it did not work. Is there a good strategy for solving simultaneous system of congruences when their modulus consist of the same multiples?
$910 = 2 \times 5 \times 7 \times 13$ and $1001 = 7 \times 11 \times 13$. Therefore, from the first statement we have:
$$x \equiv 0 \pmod 2, \ \ 2 \pmod 5, \ \ 2 \pmod 7, \ \ 2 \pmod {13}$$
and from the second statement we have: $$x \equiv 2 \pmod {7}, \ \ 5 \pmod {11}, \ \ 2 \pmod {13}$$
so now you can use the Chinese remainder theorem on:
$$x \equiv 0 \pmod 2$$ $$x \equiv 2 \pmod 5$$ $$x \equiv 2 \pmod 7$$ $$x \equiv 5 \pmod {11}$$ $$x \equiv 2 \pmod {13}$$