-1

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
  • 508
  • 4
  • 12
Joann Sammut
  • 61
  • 4
  • 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

3

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

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
  • 1,821
  • 2
  • 14
  • 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