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Solve the following linear congruences:

(i) $23x \equiv 16$ mod $107$

(ii) $234x \equiv 20$ mod $366$

(iii) $234x \equiv 6$ mod $366$.

I am trying to solve these through the use of the Euclidean algorithm but getting no where thus far. Happy to attempt the rest if someone can show me how to do at least one of the above.

Mondli.K
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  • $23x\equiv 16\pmod {107}$ is another way of saying $23x-16=107k$ for some integer $k$.Now can you use the Extended Euclidean algorithm to find $x$ and $k$? – rah4927 Nov 23 '14 at 13:12
  • @Mondli.K, You may try : http://math.stackexchange.com/questions/407478/solving-a-linear-congruence/407486#407486 – lab bhattacharjee Nov 23 '14 at 14:05

1 Answers1

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you can write $x\equiv\frac{16}{23}\equiv \frac{16+107}{23}=\frac{123}{23}\equiv \frac{123+107}{23}\equiv\frac{230}{23}\equiv 10 \mod 107$
the second equation has no solution, why?

Dr. Sonnhard Graubner
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