Question-214916

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

$$\mathrm{2017}^{\mathrm{2017}^{\mathrm{2017}} } \equiv\mathrm{1}^{\mathrm{2017}^{\mathrm{2017}} } =\mathrm{1}\left({mod}\:\mathrm{16}\right) \\ $$$$\mathrm{2017}^{\mathrm{2017}^{\mathrm{2017}} } \equiv\mathrm{142}^{\mathrm{2017}^{\mathrm{2017}} } \left({mod}\:\mathrm{625}\right) \\ $$$$\phi\left(\mathrm{625}\right)=\mathrm{500} \\ $$$$\mathrm{2017}^{\mathrm{2017}} \equiv\mathrm{52}\left({mod}\:\mathrm{125}\right);\:\mathrm{2017}^{\mathrm{2017}} \equiv\mathrm{1}\left({mod}\:\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{2017}^{\mathrm{2017}} =\mathrm{125}{a}+\mathrm{52}=\mathrm{4}{b}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{125}{a}+\mathrm{52}\equiv\mathrm{1}\left({mod}\:\mathrm{4}\right)\Rightarrow{a}\equiv\mathrm{1}\left({mod}\:\mathrm{4}\right)\Rightarrow{a}=\mathrm{4}{c}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{2017}^{\mathrm{2017}} =\mathrm{125}\left(\mathrm{4}{c}+\mathrm{1}\right)+\mathrm{52}=\mathrm{500}{c}+\mathrm{177} \\ $$$$\Rightarrow\mathrm{2017}^{\mathrm{2017}^{\mathrm{2017}} } \equiv\mathrm{142}^{\mathrm{177}} \equiv\mathrm{27}\left({mod}\:\mathrm{625}\right) \\ $$$$\Rightarrow\mathrm{2017}^{\mathrm{2017}^{\mathrm{2017}} } =\mathrm{625}{d}+\mathrm{27}=\mathrm{16}{e}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{625}{d}+\mathrm{27}\equiv\mathrm{1}\left({mod}\:\mathrm{16}\right)\Rightarrow{d}\equiv\mathrm{6}\left({mod}\mathrm{16}\right) \\ $$$$\Rightarrow{d}=\mathrm{16}{f}+\mathrm{6}\Rightarrow\mathrm{625}{d}+\mathrm{27}=\mathrm{625}\left(\mathrm{16}{f}+\mathrm{6}\right)+\mathrm{27} \\ $$$$\Rightarrow\mathrm{2017}^{\mathrm{2017}^{\mathrm{2017}} } \equiv\mathrm{10000}{f}+\mathrm{3777} \\ $$$$\Rightarrow{Last}\:\mathrm{4}\:{digits}\:{of}\:\mathrm{2017}^{\mathrm{2017}^{\mathrm{2017}} } =\mathrm{3777} \\ $$

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

Commented by MathematicalUser2357 last updated on 31/Dec/24

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

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