Question Number 227371 by Math1 last updated on 18/Jan/26
$$\mathrm{3}^{\mathrm{444}} \:\:+\:\:\mathrm{4}^{\mathrm{333}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\mathrm{dividing}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{by}\:\mathrm{7} \\ $$
Answered by A5T last updated on 19/Jan/26
$$\phi\left(\mathrm{7}\right)=\mathrm{6} \\ $$$$\mathrm{444}=\mathrm{6q}\:\mathrm{and}\:\mathrm{333}=\mathrm{6k}+\mathrm{3} \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{444}} +\mathrm{4}^{\mathrm{333}} =\left(\mathrm{3}^{\mathrm{6}} \right)^{\mathrm{q}} +\left(\mathrm{4}^{\mathrm{6}} \right)^{\mathrm{k}} \mathrm{4}^{\mathrm{3}} \equiv\mathrm{1}+\mathrm{1}\centerdot\mathrm{4}^{\mathrm{3}} \equiv\mathrm{65}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$