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 10062 by ridwan balatif last updated on 22/Jan/17

Commented by ridwan balatif last updated on 22/Jan/17

find the remainder of the division 20^(2017) +1^(2017) +17^(2017) +72^(2017) by 2017

$$\mathrm{find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{of}\:\mathrm{the}\:\mathrm{division}\:\mathrm{20}^{\mathrm{2017}} +\mathrm{1}^{\mathrm{2017}} +\mathrm{17}^{\mathrm{2017}} +\mathrm{72}^{\mathrm{2017}} \:\mathrm{by}\:\mathrm{2017} \\ $$

Answered by nume1114 last updated on 23/Jan/17

from Fermat′s little theolem: a^(p−1) ≡1 (mod p) where a∈Z,p is prime. ∴a^p ≡a (mod p) substitute: a=20,1,17,72 and p=2017 we get 20^(2017) +1^(2017) +17^(2017) +72^(2017) ≡20+1+17+72(mod 2017) =110

$$\mathrm{from}\:\mathrm{Fermat}'\mathrm{s}\:\mathrm{little}\:\mathrm{theolem}: \\ $$$${a}^{{p}−\mathrm{1}} \equiv\mathrm{1}\:\left({mod}\:{p}\right) \\ $$$$\mathrm{where}\:{a}\in\mathbb{Z},{p}\:\mathrm{is}\:\mathrm{prime}. \\ $$$$\therefore{a}^{{p}} \equiv{a}\:\left({mod}\:{p}\right) \\ $$$$\mathrm{substitute}: \\ $$$${a}=\mathrm{20},\mathrm{1},\mathrm{17},\mathrm{72}\:\mathrm{and}\:{p}=\mathrm{2017} \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$$\:\:\:\:\mathrm{20}^{\mathrm{2017}} +\mathrm{1}^{\mathrm{2017}} +\mathrm{17}^{\mathrm{2017}} +\mathrm{72}^{\mathrm{2017}} \\ $$$$\equiv\mathrm{20}+\mathrm{1}+\mathrm{17}+\mathrm{72}\left({mod}\:\mathrm{2017}\right) \\ $$$$=\mathrm{110} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com