Solve $x\equiv 1\pmod2$, $x\equiv 2\pmod3$, $x\equiv 3\pmod4$, $x\equiv 4\pmod5$, $x\equiv 5\pmod6$ and $x\equiv 0\pmod7$ $$\begin{align*}
x&\equiv 1\pmod2\\
x&\equiv 2\pmod3\\
x&\equiv 3\pmod4\\
x& \equiv 4\pmod5\\
x&\equiv 5\pmod6\\
x&\equiv 0\pmod7\\
\end{align*}$$
So the solution says we can eliminate $x\equiv 5(\bmod6)$ because the first two cases cover it, but I don't really know how it does. How do we solve it in cases like this where the moduli are not mutually relatively prime.

Yes we can eliminate it indeed


*

*$x\equiv 1 \pmod 2 \implies x=2k+1$

*$x\equiv 2 \pmod 3 \implies 2k+1\equiv 2 \pmod 3\implies k\equiv 2  \pmod 3\implies k=3h+2$
that is
$$x=6k+5 \implies x\equiv 5\pmod 6$$
Note that we can also eliminate $x\equiv 1 \pmod 2$ since we have that $x\equiv 3 \pmod 4$.
Then we can use CRT to solve the system:


*

*$x\equiv 2 \pmod 3$

*$x\equiv 3 \pmod 4$

*$x\equiv 4 \pmod 5$

*$x\equiv 0 \pmod 7$
