The following was confusing me, so any help/explanations would be greatly appreciated!

Find all incongruent solutions to the following linear congruence:

21x ≡ 14 (mod 91)

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- Sep 27th 2007, 04:03 PMclockinglyLinear Congruence
The following was confusing me, so any help/explanations would be greatly appreciated!

Find all incongruent solutions to the following linear congruence:

21x ≡ 14 (mod 91) - Sep 29th 2007, 04:12 PMThePerfectHacker
$\displaystyle \gcd(21,91)=7$ and $\displaystyle 7|14$ so there shall be $\displaystyle 7$ solutions in each congruence class. By inspection we see that $\displaystyle x\equiv 5$ is one equivalence class which solves this. By theorem it means $\displaystyle 5 + \frac{91}{7}t$ for $\displaystyle t=0,1,...,6$ shall all be solutions.

- Sep 30th 2007, 07:07 AMtopsquark
- Sep 30th 2007, 08:05 AMThePerfectHacker
Yes.

For example $\displaystyle x\equiv 2 (\bmod 4)$ the solutions $\displaystyle x=2,6$ are really the same ("congruent") solutions because they are contained in the same equivalence class (equivalence class mod $\displaystyle 4$*)

*)The congruence classes are $\displaystyle \{...,-4,0,4,...\} , \{...,-3,1,5,...\}, \{...,-2,2,6,...\}, \{...,-1,3,7,...\}$. So $\displaystyle x=2,6$ are in the same equivalence class. - Jun 3rd 2008, 10:04 AMduggaboyhmm
so if showing the incongruences would it be x=-12=14 (mod13)??