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I have the following congruence equation system:

$$ \left\{ \begin{array}{c} x \equiv 7 \pmod{7} \\ x \equiv 4 \pmod{12} \\ x \equiv 16 \pmod{21} \\\end{array} \right. $$

I understand that:

$$x\equiv a (mod w) \wedge \ x \equiv b (mod k)$$ has a solution iff $$gcd(w,k)| (b-a)$$

So I operated and found:

$$gcd(7,12) = 1 | (7-4) \rightarrow 1 | 3 : true$$ $$gcd(12,21) = 3 | (16-4) \rightarrow 3 | 12 : true$$ $$gcd(7,21) = 7 | (16-7) \rightarrow 9 | 7 : false$$

And concluded that the system does not have a solution, however, it appears to be a wrong procedure according to my teacher, and I would like to know the correct approach.

Thanks in advance.

Bill Dubuque
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Boorger
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    The first congruence means that $x$ is divisible by $7$ , but the third means $x=21k+16$ for some integer $k$ and this is not divisible by $7$. Hence the system has no solution. – Peter May 10 '22 at 21:27
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    Correct, reducing the first and third mod their moduli gcd $= 7,$ yields $\bmod 7!:,\ 7\equiv x\equiv 16,,$ which is a contradiction, cf. linked CRT solvability criterion. – Bill Dubuque May 11 '22 at 01:42

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The third congruence gives you that $x=21k+16$ for some $k\in\mathbb{Z}$. This also then means that

$$x=21k+16\equiv2\pmod{7},$$

which contradicts the first congruence. Thus the system has no solution.

Lorago
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