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In the solution to a

$ax \equiv b \pmod m$

$71x \equiv 1 \pmod{771} \implies x \equiv -76 \pmod{771}$

and

$71x \equiv 1 \pmod{77} \implies x \equiv -13 \pmod{77}$

I'm unsure of how these congruences hold, how does one divide modulo $771$?

Jyrki Lahtonen
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Asad3000
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  • My apologies, I'll learn how to format properly when next I ask a question, however I hope the current format isn't too difficult on the eyes – Asad3000 May 08 '17 at 08:52

2 Answers2

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It's worth to notice that $1\equiv -770\pmod{771}$ and $71\equiv -700\pmod{771}$ then you can divide by $70$ to get $$-10x\equiv -11\pmod{771}$$ which is equivalent to $$10x\equiv 11\pmod{771}$$ and since $11\equiv -760\pmod{771}$ we have $$10x\equiv-760\pmod{771}$$ dividing by $10$ we get $$x\equiv-76\pmod{771}$$

kingW3
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  • Thanks very much, you've clarified this instance, however what about with smaller modulo, e.g. 77x = 17mod33? – Asad3000 May 08 '17 at 09:10
  • I've got it down as => 7x = 17mod3 => 7x = 2mod3 so far. – Asad3000 May 08 '17 at 09:13
  • You can't really solve this one as $77\equiv 11\pmod{33}$ and $11x\equiv 11,22,0\pmod{33}$.This is because $\gcd(11,33)\neq 1$ so this example is doesn't have a solution so you can't find it though in general you could solve $11x\equiv 22\pmod{33}$ for example by dividing everything by $11$ to get $x\equiv 2\pmod{3}$ – kingW3 May 08 '17 at 09:20
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The calculations are easy using a fractional form of the Extended Euclidean Algorithm.

${\rm mod}\ 77\!:\, \ \dfrac{0}{77}\overset{\large\frown}\equiv \dfrac{1}{71}\overset{\large\frown}\equiv\dfrac{-1}6\overset{\large\frown}\equiv\dfrac{13}{-1}$

Equivalently, without fractions

$\begin{array}{rrl} [1]\!:\!\!\!& 77\,x\!\!\!&\equiv\ 0\\ [2]\!:\!\!\!& 71\,x\!\!\!&\equiv\ 1\\ [1]\ \ -\ \ [2]=:[3]\!:\!\!\!& 6\,x\!\!\!&\equiv -1\\ [2]-12[3]=:[4]\!:\!\!\!& -x\!\!\! &\equiv 13\\ \end{array}$

Similarly for the first inversion

${\rm mod}\ 771\!:\,\ \dfrac{0}{771}\overset{\large\frown}\equiv \dfrac{1}{71}\overset{\large\frown}\equiv \dfrac{-11}{-10}\overset{\large\frown}\equiv \dfrac{-76}1$

Bill Dubuque
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