Question-214916

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

Answered by A5T last updated on 23/Dec/24

$${2017}^{{2017}^{2017}}\equiv 1^{{2017}^{2017}}=1\left(mod16\right)$$$${2017}^{{2017}^{2017}}\equiv {142}^{{2017}^{2017}}\left(mod625\right)$$$$\varphi \left(625\right)=500$$$${2017}^{2017}\equiv 52\left(mod125\right);{2017}^{2017}\equiv 1\left(mod4\right)$$$$\Rightarrow {2017}^{2017}=125a+52=4b+1$$$$\Rightarrow 125a+52\equiv 1\left(mod4\right)\Rightarrow a\equiv 1\left(mod4\right)\Rightarrow a=4c+1$$$$\Rightarrow {2017}^{2017}=125(4c+1)+52=500c+177$$$$\Rightarrow {2017}^{{2017}^{2017}}\equiv {142}^{177}\equiv 27\left(mod625\right)$$$$\Rightarrow {2017}^{{2017}^{2017}}=625d+27=16e+1$$$$\Rightarrow 625d+27\equiv 1\left(mod16\right)\Rightarrow d\equiv 6\left(mod16\right)$$$$\Rightarrow d=16f+6\Rightarrow 625d+27=625(16f+6)+27$$$$\Rightarrow {2017}^{{2017}^{2017}}\equiv 10000f+3777$$$$\Rightarrow Last4digitsof{2017}^{{2017}^{2017}}=3777$$

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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