If $a = r \pmod b$ when $0 \le r < b$, then $\gcd(a, b) = \gcd (b, r)$

Asked Sep 09 '13 at 18:23

Active Jan 17 '14 at 03:07

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If $a = r \pmod b$ when $0 \le r < b$, then $\gcd(a, b) = \gcd (b, r)$.

I do not understand why this is, can somebody explain?

I've also looked over this thread: Why is $\gcd(a,b)=\gcd(b,r)$ when $a = qb + r$? , but I still can't seem to see how that applies to $a = r \pmod b$.

edited Apr 13 '17 at 12:20

Community

asked Sep 09 '13 at 18:23

Leon Helmsley

2

$ a = r (mod\ b) $ is same as $ a = qb + r $. – hjpotter92 Sep 09 '13 at 18:26
How are they equal? – Leon Helmsley Sep 09 '13 at 18:28
Please read this wiki entry thoroughly. – hjpotter92 Sep 09 '13 at 18:30

0 Answers0