$\frac{723}{104} (\bmod{61})$?

Question

Am I allowed to do modular arithmetic in the sense that $\frac{723}{104}\equiv \frac{723}{-18} (\bmod{61})$?

And in general how would you attack this problem?

You are supposed to compute $\frac{723}{104}$ modulo $61$, which by the very definition is a problem in modular arithmetic. So the answer to your first quation is "yes". As a first step you could replace the appearing integers by integers smaller than $61$. To systematically find $\frac{1}{104}$ modulo $61$ you could try to apply the extended Euclidean algorithm instead of just trial and error. — Hagen Knaf, Nov 19 '16 at 14:48

score 3 · Accepted Answer · edited Apr 13 '17 at 12:19

3

Yes, continuing we have $\ {\rm mod}\ 61\!:\,\ \dfrac{723}{104}\equiv \dfrac{-9}{-18}\equiv \dfrac{1}2\equiv\dfrac{62}{2}\equiv 31$

Generally we can compute modular inverses using the Extended Euclidean Algorithm (which can be implemented very conveniently by hand). But that algorithm may be a bit overkill for small numbers, where it is often simpler to employ Gauss's Algorithm and some twiddling as above.

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

edited Apr 13 '17 at 12:19

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answered Nov 19 '16 at 15:02

Bill Dubuque

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I'm starting to understand how efficient the Gauss Algorithm is, thanks a lot for introducing it to me. – Postskjerm Nov 20 '16 at 15:25

$\frac{723}{104} (\bmod{61})$?

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