Question

Answer 1

\(0,7.\left(2013^{2013}+2017^{2017}\right)\)=\(\dfrac{7}{10}.\left(2013^{2013}+2017^{2017}\right)\)

*Vì 7 không chia hết cho 10 nên ta sẽ chứng minh (2013²⁰¹³+2017²⁰¹⁷) chia hết cho 10, tức là (2013²⁰¹³+2017²⁰¹⁷)=\(\overline{...0}\)

*Ta có: 2013^4k=\(\overline{...1}\) ; 2013^4k+1=\(\overline{...3}\) ; 2013^4k+2=\(\overline{...9}\) ; 2013^4k+3=\(\overline{...7}\)

- Mà 2013 chia hết cho 4 dư 1 nên 2013²⁰¹³=\(\overline{...3}\)

*Ta có: 2017^4h=\(\overline{...1}\) ; 2017^4h+1=\(\overline{...7}\) ; 2017^4h+2=\(\overline{...9}\) ; 2017^4h+3=\(\overline{...3}\).

- Mà 2017 chia 4 dư 1 nên 2017²⁰¹⁷=\(\overline{...7}\)

*2013²⁰¹³+2017²⁰¹⁷=\(\overline{...3}\)+\(\overline{...7}\)=\(\overline{...0}\) ⋮ 10

- Vậy \(0,7.\left(2013^{2013}+2017^{2017}\right)\) là số tự nhiên.