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How to solve this linear congruence equation?

How to solve $6x \equiv 5 \mod 14$?

James
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2 Answers2

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It can't have a solution since $6$ and the modulus are even, and $5$ is odd.

Bernard
  • 175,478
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By definition, $\ 6x\equiv 5\pmod{14}\iff 6x+14y\, =\, 5\ $ for some integer $\,y$.

Deduce a contradiction by comparing the parity of both sides of the prior equaton.

Remark $\ $ Generally $\ ax\equiv b\pmod n\iff ax\!+\!ny\,=\, b\,$ for some integer $\,b.\,$
So $\, \gcd(a,n)\,$ divides $\,ax\!+\!ny = b,\,$ i.e. $\ \gcd(a,n)\mid b\,$ is a necessary condition for solvability. By Bezout's identity for the gcd this condition is also sufficient for solvability (hint: scale the Bezout identity for the gcd).

Bill Dubuque
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