On my attempt :

1) $a \equiv b\;(\operatorname{mod} m )$

2) $c \equiv d\;(\operatorname{mod} m )$

2) Implies that $kc \equiv kd\;(\operatorname{mod} m )$

so $a-kc \equiv b-kd\;(\operatorname{mod} m ),$

and then I got stuck.

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    $\begingroup$ This question as currently written seems ill-posed. What is the relationship between $a$ and $c$ and $k$? If $a$ and $c$ and $k$ are arbitrary values there is no reason for this to be true. $\endgroup$ – Mike May 1 '18 at 15:48
  • $\begingroup$ I think the question would be wrong please check your question .this equation is not possible for all values of k ,I post my answer please read it. $\endgroup$ – NEW SUN May 13 '18 at 2:38

This problem cannot be possible for all values of k. We can easily prove it . let

$5\equiv2(mod3)$ and $10\equiv1(mod3) .

Then $5-10(k)\equiv0(mod3)\implies2-k\equiv0(mod3).$

If $k=1,3,4,...$ is contradict the equation. So it is not possible for all values of k so we can find the values of k for the given values of $a,b,c,d $and $m $that can solve for above equator


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