Tuesday, 22 November 2016

linear algebra - Find all solutions of the following congurence: $52x equiv 15 (text{ mod } 91)$





Find all solutions $x \in \mathbb{Z}_{m}$ of the following congruence,
whereby $m$ is the modulus. If there isn't a solution, state why.
$$52x \equiv 15 (\text{ mod } 91)$$




I'm not sure how to solve it because if we look at $52$ and $91$, we see that they aren't coprime. So we cannot use euclidean algorithm to continue because we haven't got $\text{gcd }(52,91)=1$.



Does that mean that there won't exist a solution? Or there is another way of solving it?



Answer



Hints:



Fill in details



$$52x=15+91k\;,\;\;k\in\Bbb Z\implies15=13(4x-7k)$$



So how many solutions can you find?


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