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I'm trying to solve the problem that given any $x \in \mathbb{Z}/3705\mathbb{Z}$, I want to find all $x$ such that when mapped under the Chinese Remainder Theorem, is mapped to $(\bar{4},\bar{7})$ in $\mathbb{Z}/57\mathbb{Z} \times \mathbb{Z}/65\mathbb{Z}$.

Now I tried solving the system of congruences, $$x \equiv 4 \ (\text{mod} \ 57)$$ $$x \equiv 7 \ (\text{mod} \ 65)$$

Then I was left with, $$57k \equiv 3 \ (\text{mod} \ 65)$$ where $x=4+57k$. Now I could solve this by repeatedly adding (or subtracting) $65$ onto $3$, until I get a multiple of $57$. However when I typed it into Wolfram, the first solution that comes up is when I add or subtract $78$ and $38$ times respectively, which isn't exactly efficient in a test scenario. So I was wondering if there is some faster method to solve this question/system of congruences. Or some systematic way to find the number of times I need to add or subtract $65$.

ASP
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  • The linked dupes describe most all the known methods for computing modular inverses and fractions with many worked examples (see also their sidebar "Linked" questions for hundreds more). – Bill Dubuque May 21 '20 at 03:13
  • e.g. $,\bmod 65!:\ k\equiv \dfrac{3}{57}\equiv \dfrac{3}{-8}\equiv \dfrac{24}{-64}\equiv \dfrac{24}1,$ by Gauss's algorithm – Bill Dubuque May 21 '20 at 03:24
  • You can easily compute the exact multiple of $65$ needed so that $,57\mid 3+65k,,$ i.e. so the quotient $,\dfrac{3+65k}{57},$ is exact - see inverse reciprocity in the links (it is equivalent to two steps of the extended Euclidean algorithm). – Bill Dubuque May 21 '20 at 03:29

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