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Suppose I want to find the inverse of $7^{-1} \equiv 1 \mod 19 $, is there a quick way to do instead of extended euclidean algorithm? (Assume the numbers are not large).

Thank you

Matthew Conroy
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Katherine
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  • http://math.stackexchange.com/questions/407478/solving-a-linear-congruence – lab bhattacharjee Dec 10 '16 at 04:20
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    With small numbers, if you are reasonably quick at calculation , you will realize that you only need to recite $7$ multiples of $19$, and add one to each, and check which is divisible by $7$. It took me $8$ seconds to realize $7*11 \equiv 1 \mod 19$, it could take you a shorter time. – Sarvesh Ravichandran Iyer Dec 10 '16 at 04:21
  • $7\times 8 =56 \equiv -1 \bmod 19$ so $7^{-1}\equiv -8 \equiv 11 \bmod 19$ – Joffan Dec 10 '16 at 04:22

3 Answers3

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${\rm mod}\ 19\!:\,\ \dfrac{1}7\equiv \dfrac{3}{21}\equiv\dfrac{22}{2}\equiv 11\ $ by Gauss's algorithm. $ $ Alternatively

${\rm mod}\ 19\!:\,\ \dfrac{1}7\equiv \dfrac{1}{-12}\equiv\dfrac{\color{#0a0}{-1}}3\,\dfrac{\color{#c00}1}4\equiv\dfrac{\color{#0a0}{18}}3\,\dfrac{\color{#c00}{20}}4\equiv 6\cdot5\equiv 11 $

The extended Euclidean algorithm is also easier in fraction form

${\rm mod}\ 19\!:\,\ \dfrac{0}{19}\overset{\large\frown}\equiv\dfrac{1}7\overset{\large\frown}\equiv\dfrac{-3}{-2}\overset{\large\frown}\equiv\dfrac{11}1\quad $ [see this answer for the algorithm]

Beware $\ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.

Bill Dubuque
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Yeah, the way most people do it in competitive programming is as follows:

Suppose you have a number $k$ that is coprime to $n$ and you want to find the inverse of $k$ mod $n$. Then it is just $k^{\varphi(n)-1}$.

You can calculate this really fast using exponentiation by squaring.

Asinomás
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I sometimes look at the two multiples that straddle the modular base.

$\color{blue}{2}\cdot 7 = 14 \equiv \color{green}{-5} \bmod 19 \\ \color{orange}{3} \cdot 7 = 21 \equiv \color{red}{2} \bmod 19 $

And then combine those to get the inverse:

$1\cdot \color{green}{-5} + 3\cdot \color{red}{2} = 1 \\ \implies (1\cdot \color{blue}{2}+ 3\cdot \color{orange}{3}) \cdot 7 \equiv 1 \bmod 19 \\ \implies 7^{-1} \equiv 11 \bmod 19$

Joffan
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