$X \equiv 1 \pmod{20} \land X \equiv 1 \pmod{22} \iff X \equiv 1 \pmod{\operatorname{lcm}(20,22)}$

Asked Jun 27 '23 at 10:20

Active Jun 27 '23 at 11:38

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I dont understand how I would go about this proof. I am trying to use this to solve a CRT-problem with these two numbers, which are not coprime. I know, that you could split $20 = 2^2 \cdot 5$ and $22 = 2 \cdot 11$ in its prime factors. I know, that you could rewrite as $20 \mid X - 1$ and $22 \mid X - 1$. But I dont get it even if wikipedia and other articles act like you can see it directly. I appreciate every hint.

edited Jun 27 '23 at 11:38

Gary

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asked Jun 27 '23 at 10:20

hack03er

1

Asked very recently here... – TheSilverDoe Jun 27 '23 at 10:22
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Does this answer your question? How can I show that: $x\equiv1\pmod{20}$ and $x\equiv1\pmod{22}$ iff $x\equiv1\pmod{220}$? – lulu Jun 27 '23 at 10:29
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This is practically the definition of LCM. The first says that $20$ is a factor of $X-1$ and the second that $22$ is a factor of $X-1.$ We have a general theorem that if $a\mid c$ and $b\mid c$ then $\operatorname{lcm}(a,b)\mid c.$ – Thomas Andrews Jun 27 '23 at 10:35
Alternative (simpler, but much less elegant) approach: $$x \equiv 1 \pmod{220} \implies x \equiv 1 \pmod{20} ~~\text{and}~~ x \equiv 1 \pmod{11}.$$ For the converse, $~x \equiv 1 \pmod{20}~$ implies that the least non-negative residue of $~x, \pmod{220}~$ must be some element in $~{1,21,41,61,\cdots, 201}.$ Of these $~11~$ residues, which one(s) are congruent to $~1 \pmod{11}?$ – user2661923 Jun 27 '23 at 12:12

$X \equiv 1 \pmod{20} \land X \equiv 1 \pmod{22} \iff X \equiv 1 \pmod{\operatorname{lcm}(20,22)}$

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