Question Number 205842 by BaliramKumar last updated on 31/Mar/24
$$\frac{\mathrm{32}^{\mathrm{32}^{\mathrm{32}} } }{\mathrm{9}}\:\overset{\mathrm{R}} {\equiv}\:? \\ $$
Answered by A5T last updated on 01/Apr/24
$$\mathrm{32}^{\mathrm{32}^{\mathrm{32}} } \overset{\mathrm{9}} {\equiv}\mathrm{5}^{\mathrm{32}^{\mathrm{32}} } ;\:\mathrm{5}^{\mathrm{3}} \overset{\mathrm{9}} {\equiv}−\mathrm{1}\Rightarrow\mathrm{32}^{\mathrm{32}^{\mathrm{32}} } =\mathrm{5}^{\mathrm{3}{k}+\mathrm{1}} \equiv\left(\mathrm{5}^{\mathrm{3}} \right)^{{k}} \left(\mathrm{5}\right) \\ $$$$\equiv\left(−\mathrm{1}\right)^{{k}} \left(\mathrm{5}\right)\equiv−\mathrm{5}\left({since}\:{k}\:{is}\:{odd}\right)\overset{\mathrm{9}} {\equiv}\mathrm{4} \\ $$
Answered by A5T last updated on 01/Apr/24
$$\mathrm{32}^{\mathrm{32}^{\mathrm{32}} } \overset{\mathrm{9}} {\equiv}\mathrm{4}^{\mathrm{32}^{\mathrm{32}} } \\ $$$$\mathrm{4}^{\mathrm{3}} \equiv\mathrm{64}\overset{\mathrm{9}} {\equiv}\mathrm{1}\Rightarrow\mathrm{4}^{\mathrm{32}^{\mathrm{32}} } \equiv\mathrm{4}^{\left(−\mathrm{1}\right)^{\mathrm{32}} } \overset{\mathrm{9}} {\equiv}\mathrm{4} \\ $$