How many possibilities are there for the number of solutions of a linear congruence modulo 20

Asked Mar 07 '21 at 17:22

Active Mar 07 '21 at 17:35

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So I was given this question and I'm not very good at this if anyone could provide a little help

I know that $ax\equiv b\pmod{20}$ and by using $gcd(a,20) = 1$ unique solution and $gcd(a,20) = 2,4,5,10,20$

what do I do next? Do I test out all of the gcd?

edited Mar 07 '21 at 17:35

asked Mar 07 '21 at 17:22

beepboop

1

Please use MathJax to format your posts. To begin with, surround math expressions (including numbers) with $ signs and use _ for subscripts. $x_1$ comes out as $x_1$. – saulspatz Mar 07 '21 at 17:24
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If $gdc(a,20)\neq 1$, the equation may have no solution. For example, $2x\equiv 3 (\mod 20)$. If $1\neq c = gdc(a,20)$, then $c\mid b$ – Martín Vacas Vignolo Mar 07 '21 at 17:26
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By general theory there are no solutions if $,d = (a,20)\nmid b,,$ else there are $,d,$ solutions. What more do you need? – Bill Dubuque Mar 07 '21 at 19:01

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