Possible solutions for n * 50886 = r mod (70001)

Asked Mar 27 '23 at 23:07

Active Mar 27 '23 at 23:21

Viewed 19 times

Let n be all natural numbers, n = 1, 2, 3, ...

How to find all possible remainders in n * 50886 = r mod (70001)?

I calculated all possible values for r up to 50886.

They are: 50886, 31771, 12656, 6197, 5935, 5673, 5411, 5149, 4887, 4625, 4363, 4101, 3839, 3577, 3315, 3053, 2791, 2529, 2267, 2005, 1743, 1481, 1219, 957, 695, 433, 171, 80, 69, 58, 47, 36, 25, 14, 3, 1 and 0.

Is their a way to come up with the solution, without having to iterate over n, linearly?

edited Mar 27 '23 at 23:21

Bill Dubuque

272,048

asked Mar 27 '23 at 23:07

Johannes Heß

By the linked dupe the map $,n\to an\bmod m,$ is a bijection when $\gcd(a,m) = 1,,$ which is true above. – Bill Dubuque Mar 27 '23 at 23:20
Indeed $70001$ is a prime, so any residue $r$ can be attained by exactly one residue $n$. – hardmath Mar 27 '23 at 23:23

Possible solutions for n * 50886 = r mod (70001)

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