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Find all the solutions of each of the linear congruences below:

\begin{align} &(a) &10x &\equiv 5 \pmod{15},\\ &(b) &6x &\equiv 7 \pmod{26},\\ &(c) &7x &\equiv 8 \pmod{11}. \end{align}

I'm not entirely sure how to get these solutions by hand. I know how to prove there are solutions.

For example:

$(a) \quad\gcd(10,15)=5 $ and we know $5|5$.

From there I set $10x+15y=5$ and divide through by $5$. Leaving us with $2x+3y=1$. I know some solutions for $x$ and $y$, such as $x=-1$ and $y=1$, but that's all I have thus far.

  • 0
    Do you know that $ax\equiv b\pmod{m}$ has solutions iff $\gcd(a,m)\mid b$, and the solutions form an arithemtic progression with difference $m/g$ for $g$ total solutions?2012-09-26
  • 0
    I think that is what I was trying to show for part $(a)$. I'm just not sure where to go afterward.2012-09-26
  • 0
    See [here](http://math.stackexchange.com/questions/186674/how-to-solve-100x-19-0-pmod23/186702#186702).2012-09-26

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