Chinese Remainder Theorem with with non-pairwise coprime moduli I have a system of congruences:
$$
x\equiv 1\pmod 2 \\
x\equiv 2\pmod 3 \\
x\equiv 3\pmod 4 \\
x\equiv 4\pmod 5 \\
x\equiv 5\pmod 6 \\
x\equiv 0\pmod 7
$$
2, 3, 4, 5, 6, 7 are non-pairwise coprime, so i can't use Chinese Reminder Theorem in this case.
$x\equiv 3\pmod 4$ yields $x\equiv 1\pmod 2$ so i can just drop 3-rd equation from the system.
Using this principle i got new system:
$$
x\equiv 1\pmod 2 \\
x\equiv 2\pmod 3 \\
x\equiv 0\pmod 7 \\
$$
Solving this system o got the answer: $497 + 42k$, which is actually wrong.
Right answer i can get here: http://maciejkus.com/chinese_remainder/ , which is: $119 + 420k$
How can i solve this system right?

$$
x\equiv 1\pmod 2 \\
x\equiv 2\pmod 3 \\
x\equiv 3\pmod 4 \\
x\equiv 4\pmod 5 \\
x\equiv 5\pmod 6 \\
x\equiv 0\pmod 7
$$
$x\equiv 5\pmod 6$ if and only if $x\equiv 1\pmod 2$ and $x\equiv 2\pmod 3$. and both conditions are already on the list. This leaves us with
$$
x\equiv 1\pmod 2 \\
x\equiv 2\pmod 3 \\
x\equiv 3\pmod 4 \\
x\equiv 4\pmod 5 \\
x\equiv 0\pmod 7
$$
$x\equiv 3\pmod 4$ implies that $x\equiv 1\pmod 2$ but not vice-versa. So we keep the first and remove the second.
$$
x\equiv 2\pmod 3 \\
x\equiv 3\pmod 4 \\
x\equiv 4\pmod 5 \\
x\equiv 0\pmod 7
$$
The solution is $x = 119$
