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Question Number 214916 by Spillover last updated on 23/Dec/24

Answered by A5T last updated on 23/Dec/24

2017^(2017^(2017) ) ≡1^(2017^(2017) ) =1(mod 16) 2017^(2017^(2017) ) ≡142^(2017^(2017) ) (mod 625) φ(625)=500 2017^(2017) ≡52(mod 125); 2017^(2017) ≡1(mod 4) ⇒2017^(2017) =125a+52=4b+1 ⇒125a+52≡1(mod 4)⇒a≡1(mod 4)⇒a=4c+1 ⇒2017^(2017) =125(4c+1)+52=500c+177 ⇒2017^(2017^(2017) ) ≡142^(177) ≡27(mod 625) ⇒2017^(2017^(2017) ) =625d+27=16e+1 ⇒625d+27≡1(mod 16)⇒d≡6(mod16) ⇒d=16f+6⇒625d+27=625(16f+6)+27 ⇒2017^(2017^(2017) ) ≡10000f+3777 ⇒Last 4 digits of 2017^(2017^(2017) ) =3777

201720172017≡120172017=1(mod16)201720172017≡14220172017(mod625)ϕ(625)=50020172017≡52(mod125);20172017≡1(mod4)⇒20172017=125a+52=4b+1⇒125a+52≡1(mod4)⇒a≡1(mod4)⇒a=4c+1⇒20172017=125(4c+1)+52=500c+177⇒201720172017≡142177≡27(mod625)⇒201720172017=625d+27=16e+1⇒625d+27≡1(mod16)⇒d≡6(mod16)⇒d=16f+6⇒625d+27=625(16f+6)+27⇒201720172017≡10000f+3777⇒Last4digitsof201720172017=3777

Answered by Spillover last updated on 26/Dec/24

Notice 2017 mod φ(φ(10000)) = 2017 mod 1600 = 417 Thus 2017²⁰¹⁷ mod φ(10000) = 2017⁴¹⁷ mod 4000 = 2177 Hence 2017^(2017²⁰¹⁷) mod 10000 = 2017²¹⁷⁷ mod 10000 = 3777

Notice 2017 mod φ(φ(10000)) = 2017 mod 1600 = 417 Thus 2017²⁰¹⁷ mod φ(10000) = 2017⁴¹⁷ mod 4000 = 2177 Hence 2017^(2017²⁰¹⁷) mod 10000 = 2017²¹⁷⁷ mod 10000 = 3777

Commented by MathematicalUser2357 last updated on 31/Dec/24

He leaved with a blank comment to edit in the future!

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