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I will first let you know the problem, then I will tell you what I tried : I have to solve the congruence system,

$20x \equiv{9} \mod{30}$

$9x \equiv{12} \mod{33}$

$36x \equiv{49} \mod{60}$

With the Chinese remainder theorem, I tried to replace the equations because $30$, $33$ and $60$ have commons factors. I found prime decomposition for all modulo $30$, $33$ and $60$ and replaced those prime factor as new modulars. By simplifying I found :

$2x \equiv{0} \mod{4}$

$x \equiv{3} \mod{5}$

$9x \equiv{1} \mod{11}$

I found $x=1728$ for the new system but it does not match with the first system. Maybe there is no solution?

Bill Dubuque
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Aezir
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    Look at $36 x \equiv 49 \pmod{60}$. – Daniel Fischer Sep 24 '20 at 17:48
  • Or $20x\equiv 9\pmod {30}$ for that matter. How would you justify your simplifying? – player3236 Sep 24 '20 at 17:49
  • For exemple : $20x \equiv{9} \mod 30$ can be written as : $20x \equiv{9} \mod 2$, $20x \equiv{9} \mod 3$ and $20x \equiv{9} \mod 5$ because $gcd(2,3, 5)=1$ then we can simplify all terms – Aezir Sep 24 '20 at 17:53
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    From $20x\equiv 9 \pmod 2$ you obtain $0\equiv 1\pmod 2$, which is a contradiction. Hence there are no solutions. – player3236 Sep 24 '20 at 17:57
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    By the dupe, there exists a solution of $,ax\equiv b\pmod{! m}!\iff! \gcd(a,m)\mid b.,$ Both the 1st & 3rd congruences fail this test. In fact both are already unsolvable $!\bmod 2$, i.e. they violate parity, so they remain unsolvable mod any even modulus, by congruence persistence, i.e. if they had a solution $!\bmod 2n,$ it would persist $!\bmod 2\ \ $ – Bill Dubuque Sep 24 '20 at 21:35

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The first system of equations can have no solution. You require $36x\equiv49\mod60$. In other words, you need $36x=60k+49$, or $36x-60k=6(6x-10k)=49$, or $6x-10k=\frac{49}{6}$. But this is impossible! Assuming $x,k\in\mathbb{Z}^1$, $6x-10k\in\mathbb{Z}$, but $\frac{49}{6}=8\frac{1}{6}\not\in\mathbb{Z}$, which is a contradiction. Thus your original system of equations has no solution.

$^1$I know you didn't explicitly $\textit{say}$ that $x$ was an integer, but it's a sufficiently common assumption in these problems I felt comfortable assuming it.

Also, you have a similar problem with $20x\equiv9\mod30$. You can reach the same conclusion by rereading the first paragraph of my solution with $36\to20,49\to9,60\to30$.

Lieutenant Zipp
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