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First off: I am completely new to modular arithmetic.

I am currently trying to solve the Problem:

$x \equiv3 \pmod{19}$

$x \equiv1 \pmod{20}$

$x \equiv2 \pmod{21}$

$x \equiv1 \pmod{22}$

$x \equiv0 \pmod{23}$

I am using the Chinese remainder theorem, which states that if one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime.

I know that $x \equiv0 \pmod{23}$ should be irrelevant to solve this problem, since it cancels itself out to $0$ in the final solution step. Also the $\gcd(20,22) \neq 1$. So I am trying to find a new statement to replace $20$ and $22$. After seeing the first answer of this post, i figured the answer I am looking for might be $x \equiv 1\pmod{220}$. But I don't know how to prove it. Is there some theorem or rule that describes this problem?

In the linked post, they converted

$x \equiv 3 \pmod 4$, and $x \equiv 5 \pmod 6$, to $x \equiv 11 \pmod{12}$

Edit 1:

I know how to find the x value, given that all bases are coprime. But I don't know how to break up modular statements without coprime bases into new statements with coprime bases.

Prem
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Crango
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  • And replace (\mod nnn) with \bmod{nnn} – Anne Bauval Jun 25 '23 at 17:29
  • @RossMillikan Sorry, now it's correct. – Crango Jun 25 '23 at 17:30
  • So, is $\mathrm{lcm}(20,22)$ equal to $220$? – Geoffrey Trang Jun 25 '23 at 17:31
  • @GeoffreyTrang Yes. – Crango Jun 25 '23 at 17:32
  • @AnneBauval Sorry, new here. What do you mean with "include the statement of the crt"? – Crango Jun 25 '23 at 17:33
  • Type the statement of the Chinese remainder theorem. – Anne Bauval Jun 25 '23 at 17:33
  • @AnneBauval: That, I think, is going too far. A link (https://en.wikipedia.org/wiki/Chinese_remainder_theorem) would be sufficient. – TonyK Jun 25 '23 at 17:35
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    Let's break this down into simpler parts first. Do you know how to do the following: I/ Given $\gcd(a, b) = 1$, $ x\equiv A \pmod{a}, x \equiv B \pmod{b}$, find such an $x$. II/ Given $\gcd(a, b) = d$, $x\equiv A \pmod{a}, x \equiv B \pmod{b}$, write some modular statements which give the same solution, BUT the bases are all coprime. $\quad$ Once you understand these, the idea is to use $II$ to break up the statement into coprime bases, and then use $I$ to build up the solution from coprime bases. – Calvin Lin Jun 25 '23 at 17:35
  • The title is now wrong again. Please change the $21$ back to $22$. If the $21$ were to be left as is, then $220$ would need to be replaced with $420$, which is not what is intended. – Geoffrey Trang Jun 25 '23 at 17:41
  • @CalvinLin Edit 1. I am struggling with step 2. – Crango Jun 25 '23 at 17:45
  • @GeoffreyTrang Sorry, I am clumsy today. – Crango Jun 25 '23 at 17:45
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    @AnneBauval \pmod (for "parenthetical mod"), not \bmod ("binary mod"; that's the binary operator). – Arturo Magidin Jun 25 '23 at 18:00
  • This is a better cite. It explains what to do when the moduli are not pairwise coprime. – Arturo Magidin Jun 25 '23 at 18:03
  • (In fact, I'm tempted to close as a duplicate of that, but I'll wait a bit...) – Arturo Magidin Jun 25 '23 at 18:09
  • @ArturoMagidin Please wait a bit. The other post is super theoretical and I am more of a practical learner. – Crango Jun 25 '23 at 18:12
  • @Crango Breaking II down even further. Prove/Demonstrate the following (not just stating that you know it) IIA) Given $a= pq$ where $\gcd(p, q) = 1$ and $ x \equiv A \pmod{a}$, then what else can we say about $ x\pmod{p}, x \pmod{q}$? IIB) (which is actually I in disguise) Given $a= pq$ where $\gcd(p, q) = 1$, if we know what is $ x \pmod{p}, x\pmod{q}$, then what can we say about $ x \pmod{a}$? – Calvin Lin Jun 25 '23 at 18:59
  • See CCRT in the 1st linked dupe for the titled question, and more generally see here and here in the 2nd & 3rd dupes for CRT with noncoprime moduli. – Bill Dubuque Jun 25 '23 at 19:23
  • @Crango: No need of CRT here, nor long answers or links to more complicated alleged duplicates: by definition of the lowest common multiple, $x-1$ is divisible by $20$ and by $22$ iff it is divisible by $\operatorname{lcm}(20,22)=220.$ – Anne Bauval Jun 26 '23 at 05:18

2 Answers2

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(A) When we have $x \equiv 1 \mod 220$ , it is Equivalent to :
$x=220X+1$

That gives :
$x=22(10X)+1$ which is $x \equiv 1 \mod 22$
$x=20(11X)+1$ which is $x \equiv 1 \mod 20$

(B) When we have $x \equiv 1 \mod 22$ & $x \equiv 1 \mod 20$ , it is Equivalent to :
$x=22Y+1$ & $x=20Z+1$
Which gives :
$x=22Y+1=x=20Z+1$
Hence , $22Y=20Z$ , $11Y=10Z$
Naturally , $Y$ must have factor $10$ , while $Z$ must have factor $11$.
$Y=10y$ & $Z=11z$

Hence , we get :
$x=22(10y)+1$ & $x=20(11z)+1$
Hence $z=220y+1$ & $x=220z+1$
Which is :
$x \equiv 1 \mod 220$

DONE

I am listing out the long calculations , because OP is new to Modular Arithmetic , though it is not necessary : When OP gets comfortable with Modular Arithmetic , it can be simplified.

Prem
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HINT.- $x-1$ divisible by $220=2^2\cdot5\cdot11$ implies that it is divisible by $2^2\cdot5=20$ and by $2\cdot11=22$.

Reciprocally $x-1$ divisible by $2^2\cdot5=20$ and by $22$ implies that it is divisible by $22$ and by $10$ so it is divisible by $22\cdot 10=220$.

Piquito
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    Not sure why this was Downvoted , It is Essentially Correct , though there is a small error here : "divisible by 22 and by 10 so it is divisible by 220" should be "110" , though that Intermediate Part in unnecessary & should be eliminated. – Prem Jun 25 '23 at 19:07
  • I didn't downvote this. – Crango Jun 25 '23 at 19:18
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jun 25 '23 at 19:19
  • @Bill Dubuque All right sir. (What happen to me is that I need to write first to be sure my wording is not totally bad because of my deficient English; is hard to me writting each answer). Besides I can't avoid wishing to help beginners which is my main purpose here in MSE (I am 84 almost 85 years old man). – Piquito Jun 27 '23 at 13:36
  • We all want to help others, but it is important to do that in a way that keeps the site healthy. When there are too many duplicates of common questions it makes it difficult if not impossible for questioners to locate optimal answers by search (which is the primary purpose of the site - to be an (easily) searchable library of answers to questions). Btw, that is not my downvote. – Bill Dubuque Jun 27 '23 at 14:55