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I am revising for my exam but could not figure a way to solve the linear congruence

$$ 57x \equiv -3 \pmod{50}.$$

I have done $-3 = 50 × (-1) + 47$ but don't think i am on right track.

May someone give me a hint please.

Thanks

Adler Marques
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Joann Sammut
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  • Hint: the coef of $x$ (which we reduce to $57\equiv 7)$ can be reduced further by scaling the equation by $k$ such that $,7k$ is very close to $50$. See Gauss's Algorithm in the linked dupes. – Bill Dubuque Feb 04 '21 at 21:39

1 Answers1

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$$ 57x \equiv 7x \pmod {50}$$

so we just need to find inverse of $7$ modulo $50$, and then multiply it by $-3$.

Observe that $7 \cdot 7 = 49 \equiv -1 \pmod{50}$, so $7^{-1} \equiv -7 \pmod{50}$. Therefore

$$x \equiv (-3)\cdot 7^{-1} \equiv (-3)\cdot(-7) =21 \pmod{50}$$

radekzak
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  • You pulled inverse of $7$ out of a hat - which is good for magic - but not good for math. For mathematical methods see the linked dupes (and hundreds of others - this is a FAQ). – Bill Dubuque Feb 04 '21 at 21:15