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Question-211632




Question Number 211632 by sonukgindia last updated on 15/Sep/24
Answered by A5T last updated on 15/Sep/24
φ(1000)=400  2024^(2024) ≡0(mod 16);2024^(2024) ≡1(mod 25)  ⇒2024^(2024) =25q+1≡(0 mod 16)⇒q≡7(mod16)  ⇒2024^(2024) =25(16k+7)+1=400k+176≡176(mod 400)  ⇒2024^(2024^(2024) ) ≡24^(176) (mod 1000)≡(24^(11) )^(16) ≡24^(16)   ≡24^(11) ×24^5 ≡24^6 ≡976(mod 1000)⇒?=976
$$\phi\left(\mathrm{1000}\right)=\mathrm{400} \\ $$$$\mathrm{2024}^{\mathrm{2024}} \equiv\mathrm{0}\left({mod}\:\mathrm{16}\right);\mathrm{2024}^{\mathrm{2024}} \equiv\mathrm{1}\left({mod}\:\mathrm{25}\right) \\ $$$$\Rightarrow\mathrm{2024}^{\mathrm{2024}} =\mathrm{25}{q}+\mathrm{1}\equiv\left(\mathrm{0}\:{mod}\:\mathrm{16}\right)\Rightarrow{q}\equiv\mathrm{7}\left({mod}\mathrm{16}\right) \\ $$$$\Rightarrow\mathrm{2024}^{\mathrm{2024}} =\mathrm{25}\left(\mathrm{16}{k}+\mathrm{7}\right)+\mathrm{1}=\mathrm{400}{k}+\mathrm{176}\equiv\mathrm{176}\left({mod}\:\mathrm{400}\right) \\ $$$$\Rightarrow\mathrm{2024}^{\mathrm{2024}^{\mathrm{2024}} } \equiv\mathrm{24}^{\mathrm{176}} \left({mod}\:\mathrm{1000}\right)\equiv\left(\mathrm{24}^{\mathrm{11}} \right)^{\mathrm{16}} \equiv\mathrm{24}^{\mathrm{16}} \\ $$$$\equiv\mathrm{24}^{\mathrm{11}} ×\mathrm{24}^{\mathrm{5}} \equiv\mathrm{24}^{\mathrm{6}} \equiv\mathrm{976}\left({mod}\:\mathrm{1000}\right)\Rightarrow?=\mathrm{976} \\ $$

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