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I mean, if you have multiples equations and all modulus are co-primes you can 'merge' them in just one equation.

But can you do it reversed? Something like:

(a mod p*q) p and q co-prime

And then create the equations (a mod p) and (a mod q). I've seen that:

18≡18(mod 35) but

18≡4(mod 7) and 18≡3(mod 5)

But I've seen other questions where it was told that you are able to do it.

Idhrenniel
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  • Yes. The Chinese remainder theorem establishes a bijection between residue classes $\pmod {pq}$ which are prime to $pq$, and pairs of classes $\pmod p, \pmod q$ both prime to their respective modulus. – lulu Jun 06 '19 at 18:11
  • Ok, but what about the example i proposed? – Idhrenniel Jun 06 '19 at 18:14
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    I don't see the problem. $18$ is a good residue class $\pmod {5\times 7}$. We check that $\gcd(18,5\times 7)=1$, so that checks. So the CRT should give us a pair of residue classes $a\pmod 5$ and $b\mod 7$ which corresponds to it. We confirm that $(a,b)=(3,4)$ so we're good. What's the problem? – lulu Jun 06 '19 at 18:25
  • Combinatorially, this means that the number of classes $\pmod {35}$ which are prime to $35$ should be the product of the number of classes $\pmod 5$ and $\pmod 7$ which are prime to their respective modulus. If you know about the Euler phi function, this just means that $\varphi(35)=\varphi(5)\times \varphi(7)$. Since the numbers are all so small you can check this by hand. – lulu Jun 06 '19 at 18:26
  • So it doesnt matter if the residues are not the same in the new both equations? I've seen that CRT had two equations with an x that should be the same – Idhrenniel Jun 06 '19 at 18:29
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    What does "the same" mean? A class $\pmod 5$ is never going to be a class $\pmod 7$. They intersect, of course. At $18$. But they are not "the same". – lulu Jun 06 '19 at 18:30

2 Answers2

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This direction of the statement is trivially true.

If $x \equiv a \pmod {pq}$, then $pq \mid (x-a)$.

So $p \mid (x-a) $ and $q \mid (x-a)$.

So, $x \equiv a \pmod p$ and $x \equiv a \pmod q$.

Bill Dubuque
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tia
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(Too long for a comment)

I think your original doubt came from a common misunderstanding about notation. $$x \equiv a \pmod {n}$$ does not mean that $x$, reduced modulo $n$, equals $a$. It means that the difference between $a$ and $x$ is a multiple of $n$, or, to look at it another way, that $a$ reduced modulo $n$ and $x$ reduced modulo $n$ are equal.

It is true that $13 \equiv -478847 \pmod {10}$, for example.

Martin Kochanski
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