Question

Cmr:(2.10²⁰¹⁷+2017)chia het cho 3

Answer 1

Có: 10 \(\equiv1\) ( mod 3 )

=> 10²⁰¹⁷ \(\equiv1\) ( mod 3)

2 . 10²⁰¹⁷ \(\equiv2\) ( mod 3 ) (1)

2017 \(\equiv1\) (mod 3 ) (2)

Từ (1) và (2) => 2 . 10²⁰¹⁷ + 2017 \(\equiv2+1\) (mod 3 )

hay 2 . 10²⁰¹⁷ + 2017 \(⋮3\left(đpcm\right)\)

 

Answer 2

Có: \(10\equiv1\left(mod3\right)\Rightarrow10^{2017}\equiv1\left(mod3\right)\)

\(2.10^{2017}\equiv2\left(mod3\right)\left(1\right)\)

\(2017\equiv1\left(mod3\right)\)

Từ (1) và (2) \(\Rightarrow2.10^{2017}+2017\equiv2+1\left(mod3\right)\)

hay 2.10²⁰¹⁷ + 2017 \(⋮\) 3 (đpcm)