I know how to find $one$ inverse via the euclidean algorithm, but I can't figure out how to find more of them.

For example:

Find an inverse $x$, of $57$ $modulo$ $100$ Or an $x$ such that $57x ≡ 1$ modulo 100

I got the answer $-7$ from the euclidean algorithm, but then the domain of x is restricted to be between $0$ and $100$. I know $93$ works, but not how I would go about finding that on paper.


  • $\begingroup$ Hint: Any two inverses will be congruent mod $100$. $\endgroup$ – Tobias Kildetoft Apr 27 '15 at 17:55
  • $\begingroup$ Hint 93 = 100-7 $\endgroup$ – Surb Apr 27 '15 at 17:56
  • $\begingroup$ you can write $\frac{1}{57} and adding or subtracting 100 to the numerator until we get a integer number $\endgroup$ – Dr. Sonnhard Graubner Apr 27 '15 at 17:59

Once you get one answer $x=-7$ then all the others can be obtained by using the fact that $x \equiv -7 \pmod{100}$. Thus every such $x$ is of the form $x=-7+100t$, where $t\in \mathbb{Z}$.

  • 2
    $\begingroup$ errrm... $\bmod 100$ cough $\endgroup$ – AlexR Apr 27 '15 at 17:57
  • $\begingroup$ @AlexR thanks for pointing the typo. $\endgroup$ – Anurag A Apr 27 '15 at 17:58
  • $\begingroup$ Man that was really simple. I feel dumb now haha. Thanks for the help. $\endgroup$ – Avernikas Apr 27 '15 at 18:03

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