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Suppose we know $x\equiv 3 \bmod 11$ and $x\equiv 7\bmod 13$ and $0<x<143$ holds then CRT gives that $x=3\times 13[13^{-1}\bmod 11] + 7\times 11[11^{-1}\bmod 13]=39\times 6 + 77\times 6=696$ gives a solution but it is not within $0$ and $143$. So we take $696\bmod 143$ and choose $124$ as solution.

My query is say instead of $3$ we choose $3+11k$ and instead of $7$ we choose $7+13\ell$ for some $k,\ell\in\Bbb Z$ can we still recover $124$?

Instead of $696$ is there a direct way to get $124$?

We have $696=124+4(143)$. In general what is the quantity that goes in instead of $4$?

Turbo
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3 Answers3

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The direct way to $124$ would be to start with $7$ and see how many $13$s to add to make it into a number that is $3\bmod 11$.

Since $13\equiv \color{red}2 \bmod 11$, we can see that we need $(3-7)/\color{red}2 = -2\equiv 9\bmod 11$ copies of $13$. $7+9\cdot 13 = 7+117 = 124$.

Joffan
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we have $$x=3+11m,x=7+13n$$ from here we get the Diophantine equation $$11m-13n=4$$ with Solutions $$m=11+13k,n=9+11k$$ plugging this in the equations for $x$ $$x=124+11\cdot 13k$$ thus we obtain $$x=124$$ for $$k=0$$

Dr. Sonnhard Graubner
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Some people do CRT theorem problems like this. If $x\equiv 3\pmod{11},$ then $x = 3+11k$. Plug that into $x \equiv 7 \pmod{13}$ to get $3+11k \equiv 7 \pmod{13}$ which reduces to

$$11k \equiv 4 \pmod{13}$$

$$-2k \equiv 4 \pmod{13}$$

$$k\equiv -2 \equiv 11 \pmod{13}$$

So $x = 3+11\cdot 11 =124.$

B. Goddard
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  • nice $\mbox{ }$ – Turbo Apr 21 '17 at 12:17
  • does this always work??? even if we do mod 13 first and 11 second and also for any pair (replace 13 and 11 by variables and 3 and 7 by variables) of modular residues – Turbo Apr 21 '17 at 12:19
  • Yes. And if you have more congruences, you can continue by plugging in $x = 124 + 142n$ into the next congruence and repeat. The official CRT requires the moduli to be relatively prime, but this method doesn't. It will find the solutions (if they exist.) – B. Goddard Apr 21 '17 at 12:25
  • Well, with all those variables flying around, it's going to be harder to solve the second congruence, for sure. And if you don't know the values of these 4 variables, it will be hard to know how big or small your answer is. – B. Goddard Apr 21 '17 at 12:41
  • In my case instead of 3 I have a $2Ak_1+3Bk_2$ and instead of $7$ I have $5Ak_1-3Bk_2$ where $A,B$ are known coprime and $k_1,k_2>0$ are unknown but smaller than $A,B$ and instead of $11$ I have $2A-B$ and instead of $13$ I have $2A+B$ ($A,B$ have to be treated as known variables). I do not see how to still express $k$ in your setting properly without introducing another $k_3$ since I know $x>0$ is of size $<AB$. I will get $k\equiv \alpha A+\beta B\bmod 2A+B$ and if I use $x=2Ak_1+3Bk_2+(2A-B)(\alpha A+\beta B)$ I will get incorrect $2Ak_1+3Bk_2+(2A-B)(\alpha Ak_1+\beta Bk_2)>AB$ – Turbo Apr 21 '17 at 12:41
  • clarified above comment – Turbo Apr 21 '17 at 12:42
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    @Robut See this answer for a proof that it always works. You can find worked examples in the list of linked posts there. – Bill Dubuque May 03 '17 at 15:52