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How to find all solutions for $12x \equiv 12 \pmod{30}$ ?

$12x \equiv 12 \pmod{30}$ gives $12x=12+30y$
$\gcd(30,12)=6$
$30=2\cdot12+6$
$12=2\cdot6$

$6=30-(2*12)$

Multiply all with 2:
$12=2\cdot30-4\cdot12$
Switch sides for easier reading:
$2\cdot30-4\cdot12=12$

I think I am right this far (correct me if I'm wrong). But how do I finish this problem? I don't understand what I'm supposed to do now.

Eric Snyder
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Bioelli
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  • You applied the Euclidean algorithm and found a solution ($x_0=-4$), but it was useless here since there is an obvious solution, $x_1=2$. Once you have a solution ($x_i=x_0$ or $x_1$), the other solutions $x$ are given by $30\mid12(x-x_i)$, i.e. $5\mid2(x-x_i)$, i.e. $5\mid x-x_i$. – Anne Bauval Sep 21 '22 at 09:34
  • Every time you use an asterisk to represent multiplication outside computer code, a mathematician somewhere dies a tiny bit inside. Please use "\cdot" ($\cdot$) or "\times" ($\times$) in MathJax. – Eric Snyder Sep 21 '22 at 10:04
  • By here, cancel $(12,30)!=!6$ to get $\bmod 30!:\ 12x\equiv 12\iff\bmod 5!:\ 2x\equiv 2\iff x\equiv 1,,$ by $(2,5)=1\Rightarrow 2$ is invertible so cancellable $!\bmod 5\ \ $ – Bill Dubuque Sep 21 '22 at 10:08

1 Answers1

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$12x=12+30y$

$2x=2+5y$

$2x\equiv 2 \space mod \space 5$

Since $(2,5)=1$

$x\equiv 1 \space mod \space 5$ or $x=1+5k$ for some integer k

Abel Wong
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