How to simplify LHS of a linear congruence?

Asked Jan 27 '22 at 14:39

Active Jan 27 '22 at 14:44

Viewed 22 times

following steps are mentioned

$$42x \equiv 1 \pmod 5\\ 2x \equiv1 \pmod 5$$

and $$ 35x \equiv 1 \pmod 6\\ 5x \equiv 1 \pmod 6 $$ so on...

How he goes from $42x$ to $2x$ and from $35x$ to $5x$?

If there is any theorem, are there any tricks / techniques to calculate those fast?

edited Jan 27 '22 at 14:44

miracle173

asked Jan 27 '22 at 14:39

Dum

3

What is $42\bmod 5$? What is $35 \bmod 6$? – Michael Albanese Jan 27 '22 at 14:43
As suggested in the comments to this video you can start with Number Theory | Congruence Modulo n -- Definition and Examples from the same person. – miracle173 Jan 27 '22 at 14:52
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By the linked Congruence Product Rule: $\bmod 5!:\ 42\equiv 2\Rightarrow 42x\equiv 2x,$ and similarly for the second one. See all the basic Congruence Rules there. – Bill Dubuque Jan 27 '22 at 15:17
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In fact $, 2 = (42 \bmod 5),$ is the remainder left on dividing $42$ by $5,,$ which can be calculated efficiently (in general) by the division algorithm. – Bill Dubuque Jan 27 '22 at 16:32

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