Question Number 226138 by AgniMath last updated on 20/Nov/25
$${What}\:{will}\:{be}\:{remainder}\:{when} \\ $$$$\left(\mathrm{9023}\right)^{\mathrm{2029}} \:{is}\:{divided}\:{by}\:\mathrm{27}? \\ $$
Commented by AgniMath last updated on 20/Nov/25
$${Anyone}\:{tell}\:{me}\:{the}\:{answer}\:{only} \\ $$
Commented by CDSSW_Legend last updated on 20/Nov/25
$$ \\ $$$${remainder}\:{of}\:\mathrm{23} \\ $$
Commented by Linton last updated on 20/Nov/25
$${Hi}\:{i}\:{would}\:{like}\:{to}\:{see}\:{full}\:{working}\:? \\ $$
Answered by A5T last updated on 20/Nov/25
$$\mathrm{9023}\:\equiv\:\mathrm{5}\:\left(\mathrm{mod}\:\mathrm{27}\right) \\ $$$$\Rightarrow\left(\mathrm{9023}\right)^{\mathrm{2029}} \:\equiv\:\mathrm{5}^{\mathrm{2029}} \:\left(\mathrm{mod}\:\mathrm{27}\right) \\ $$$$\phi\left(\mathrm{27}\right)=\mathrm{3}^{\mathrm{3}} −\mathrm{3}^{\mathrm{2}} =\mathrm{18}\:\Rightarrow\:\mathrm{5}^{\mathrm{18}} \:\equiv\:\mathrm{1}\left(\mathrm{mod}\:\mathrm{27}\right) \\ $$$$\mathrm{5}^{\mathrm{2029}} =\:\left(\mathrm{5}^{\mathrm{18}} \right)^{\mathrm{112}} \left(\mathrm{5}^{\mathrm{13}} \right)\equiv\mathrm{5}^{\mathrm{13}} =\left(\mathrm{5}^{\mathrm{2}} \right)^{\mathrm{6}} \mathrm{5}\equiv\left(−\mathrm{2}\right)^{\mathrm{6}} ×\mathrm{5} \\ $$$$\equiv\mathrm{10}×\mathrm{5}\equiv\mathrm{23}\:\Rightarrow\:\:\left(\mathrm{9023}\right)^{\mathrm{2029}} \:\equiv\:\mathrm{23}\:\left(\mathrm{mod}\:\mathrm{27}\right) \\ $$