If $a \equiv b\;(\operatorname{mod} m )$ and $c \equiv d\;(\operatorname{mod} m ),$ prove that $a - kc \equiv 0\; (\operatorname{mod} m ).$

Question

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.

Answer 1

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