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Consider following system : \begin{cases} x \equiv 2 \, \text{mod} \, 9 \\ x \equiv 3 \, \text{mod} \, 5 \\ x \equiv 8 \ \text{mod} \, 21 \end{cases} I can't apply the Chinese remainder theorem because the $\text{gcd}(21,9) \neq 1.$ So I want to factorize $21$ and $9$ in prime factors and make new equations. A theorem in my books says if the $\text{gcd}(n,m) = 1$ then $ x \equiv x' \, \text{mod}\ nm \iff \begin{cases} x \equiv x' \, \text{mod} \, n \\ x \equiv x' \, \text{mod} \, m \\ \end{cases} $. But how do I apply this with the first equation. Because $9=3*3$ and the $\text{gcd}(3,3) \neq 1$. I heard some solution who said that $x \equiv 2\,\text{mod}\, 9$ is the same as $x \equiv 2\,\text{mod}\, 3$ but I don't understand why they are the same. Thanks in advance!

Kabouter9
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1 Answers1

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Hint. Since $21=3\cdot 7$, by CRT, the system is equivalent to $$\begin{cases} x \equiv 2 \, \text{mod} \, 9 \\ x \equiv 3 \, \text{mod} \, 5 \\ x \equiv 8 \ \text{mod} \, 3\\ x \equiv 8 \ \text{mod} \, 7 \end{cases}$$ That is $$\begin{cases} x \equiv 2 \, \text{mod} \, 9 \\ x \equiv 3 \, \text{mod} \, 5 \\ x \equiv 2 \ \text{mod} \, 3\\ x \equiv 1 \ \text{mod} \, 7 \end{cases}$$ Since $x \equiv 2\ \text{mod} \, 9$ implies $x \equiv 2\ \text{mod} \, 3$,it remains to solve $$\begin{cases} x \equiv 2 \, \text{mod} \, 9 \\ x \equiv 3 \, \text{mod} \, 5 \\ x \equiv 1 \ \text{mod} \, 7 \end{cases}$$ Now apply CRT.

Robert Z
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    It is just the hint that I don't understand... Why you can replace it... – Kabouter9 Aug 15 '19 at 14:41
  • By CRT, $x \equiv 8 \ \text{mod} , (3\cdot 7)$ is equivalent to $x \equiv 8 \ \text{mod} , 3$ AND $x \equiv 8 \ \text{mod} , 7$. – Robert Z Aug 15 '19 at 14:43
  • And why is x≡2 mod9 then x≡2 mod3? – Kabouter9 Aug 15 '19 at 14:45
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    Not sure why someone downvoted this. It is a correct answer. It addresses both of the non-prime moduli by showing the system is consistent. Example of a non-consistent system: $$\begin{cases}x \equiv 2 \pmod{9} \ x \equiv 3\pmod{5} \ x\equiv 10\pmod{21}\end{cases}$$ Which is easily shown to be inconsistent since $x\equiv 2\pmod{9} \Longrightarrow x\equiv 2\pmod{3}$, but $x\equiv 10\pmod{21} \Longrightarrow x\equiv 1\pmod{3}$, a contradiction. – SlipEternal Aug 15 '19 at 14:47
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    @Kabouter9 I edited my answer. Is it clear now? – Robert Z Aug 15 '19 at 14:51
  • I'm still struggling with $x \equiv 2 , mod ,9$ implies $x \equiv 2 , mod , 3$. Intuitive it is clear, but I don't have a mathematical explanation for it. – Kabouter9 Aug 15 '19 at 14:54
  • $x \equiv 2 , mod ,9$ means that $x=9q+2=3(3q)+2$. – Robert Z Aug 15 '19 at 14:56
  • Got it! Thanks you all! – Kabouter9 Aug 15 '19 at 15:01