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Ok, I found a lot of questions asking about solving $a = b \pmod c$ where you could divide $a$ and $b$ by some $x$ where gcd$(x, c) = 1$. How do you solve when this is not the case?

Suppose I have $10 x \equiv 5 \pmod{15}$. How do I solve this? How can you solve to get a linear equation in $x$?

On inspection (and trying out values), I see that $x = 3n + 2$ is what I'm looking for. How can I get this mathematically?

And yes, this is homework, but I changed the numbers so that I could practice on the actual problem ;)

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See here. –  Mhenni Benghorbal Sep 16 '12 at 5:04

2 Answers 2

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$\begin{eqnarray}\rm{\bf Hint}\ &&\rm mod\ mc\!:\ ac\,x\equiv bc&\iff&\rm mod\ m\!:\ ax\equiv b\quad for\ \ c\ne0\\ \rm &&\rm by\ \ \ \ mc\, \:|\: \ ac\,x-bc&\iff&\rm m\ |\ ax-b\\ \rm &&\rm because\ \ \ \dfrac{ac\,x-bc}{mc}&\ \ =\ \ &\rm \dfrac{ax-b}{m} \end{eqnarray}$

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Thanks. I'll work it out. –  asymptotically Sep 17 '12 at 3:21
    
Essentially: we can cancel $\rm\,c\,$ from the congruences (or the equivalent divisibilty relations) because they are equivalent to said fraction being integral, and $\rm\,c\,$ cancels in the fraction. –  Bill Dubuque Sep 17 '12 at 3:27

Hint: In general, if $k\ne 0$, we have $$kax\equiv kb \pmod{km} \quad\text{iff}\quad ax\equiv b \pmod{m}.$$

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