Chinese Remainder Theorem with with non-pairwise coprime moduli

Question

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?

You cannot drop third, you can only drop first and second equation. — , Jan 17 '15 at 13:57
$497+42k$ is the solution of the system with the $3$ equations. Better would be the representation $35+42k$. — Peter, Jan 17 '15 at 13:57
Considering also the equation with the $mod\ 5$, we get $119+120k$ and $119$ solves the complete system already. So, we have to find the period. It is $lcm(2,3,4,5,6,7)=420$. So, the solution of the complete system is $119+420k$. — Peter, Jan 17 '15 at 14:00
It is enough to consider the equations with the moduli $3,4,5,7$ because the modulo-$4$-equation covers the modulo-$2$-equation and therefore (together wit the modulo-$3$-equation) also the modulo-$6$-equation. And now, you have pairwise coprime moduli. — Peter, Jan 17 '15 at 14:12
I think i have to add $x\equiv 4\pmod 5$ equation to the new system too. — Joey74, Jan 17 '15 at 14:14
Adding the equation with moduli 5 i got solution: $-91+210k$ — Joey74, Jan 17 '15 at 14:44
Exact Dupe of Solve $x\equiv 1\pmod2$, $x\equiv 2\pmod3$, $x\equiv 3\pmod4$, $x\equiv 4\pmod5$, $x\equiv 5\pmod6$ and $x\equiv 0\pmod7$ — Bill Dubuque, Feb 25 '23 at 00:42

score 1 · Answer 1 · answered May 05 '17 at 06:42

$$ x\equiv 1\pmod 2 \\ x\equiv 2\pmod 3 \\ x\equiv 3\pmod 4 \\ x\equiv 4\pmod 5 \\ x\equiv 5\pmod 6 \\ x\equiv 0\pmod 7 $$

$x\equiv 5\pmod 6$ 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 $x = 119$

Chinese Remainder Theorem with with non-pairwise coprime moduli

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