Question Number 208980 by lmcp1203 last updated on 30/Jun/24
$${please}\:.\:\:\:\:\:{find}\:\:\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} } {mod}\:\mathrm{23}\:\:\:\:\:\:\:\:{thanks}. \\ $$
Answered by A5T last updated on 30/Jun/24
![2^(11001^(666) ) mod(23)≡2^(11001^(666) [mod φ(23)]) ≡2(mod 23) ______________________________________ φ(23)=22; 11001^(666) (mod 22)≡1^(666) =1(mod 22)](https://www.tinkutara.com/question/Q208992.png)
$$\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} } {mod}\left(\mathrm{23}\right)\equiv\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} \left[{mod}\:\phi\left(\mathrm{23}\right)\right]} \equiv\mathrm{2}\left({mod}\:\mathrm{23}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\phi\left(\mathrm{23}\right)=\mathrm{22};\:\mathrm{11001}^{\mathrm{666}} \left({mod}\:\mathrm{22}\right)\equiv\mathrm{1}^{\mathrm{666}} =\mathrm{1}\left({mod}\:\mathrm{22}\right) \\ $$
Commented by lmcp1203 last updated on 30/Jun/24
$${thanks} \\ $$