Question

[Trần Mạnh Tiến]

CHO M =\(2^0+2^2+2^4+2^6+.+2^{2018}\)

TÌM SỐ DƯ KHI M CHIA CHO 7

HELP ME !!!

Answer 1

[Phan Minh Thiện]

\(M=2^0+2^2+2^4+2^6+2^8+...+2^{2018}\)

\(M=2^0+2^2+\left(2^4+2^6+2^8\right)+...+\left(2^{2014}+2^{2016}+2^{2018}\right)\)

\(M=1+4+2^4.\left(1+2^2+2^4\right)+...+2^{2014}.\left(1+2^2+2^4\right)\)

\(M=5+2^4.21+2^{10}.21+...+2^{2014}.21\)

\(M=5+21.\left(2^4+2^{10}+...+2^{2014}\right)\)

vì  \(21.\left(2^4+2^{10}+...+2^{2014}\right)⋮7\)

nên \(M=5+21.\left(2^4+2^{10}+...+2^{2014}\right)\)chia 7 dư 5