Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

ϕ(1000)=400202420240(mod16);202420241(mod25)20242024=25q+1(0mod16)q7(mod16)20242024=25(16k+7)+1=400k+176176(mod400)20242024202424176(mod1000)(2411)1624162411×245246976(mod1000)?=976

Terms of Service

Privacy Policy

Contact: info@tinkutara.com