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Question Number 93184 by Rio Michael last updated on 11/May/20

$$\mathrm{in}\mathrm{solving}\mathrm{the}\mathrm{linear}\mathrm{congruence}$$$$ax\equiv b\left(\mathrm{mod}n\right)\Rightarrow n\mid (ax-b)\Rightarrow ax-b=kn\iff ax-kn=b$$$$\Rightarrow \mathrm{solving}\mathrm{the}\mathrm{linear}\mathrm{diophantine}\mathrm{equation}ax-kn=b$$$$\mathrm{what}\mathrm{are}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{equation}$$$$ax-kn=b$$$$$$$$$$

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