Question

cho B= 2+2^3+2^5+...........+2^99+2^101 a) chia hết cho 3,chia hết cho 7 b)chia 5 dư 2

Answer 1

a: \(B=2\left(1+2^2+2^4\right)+2^7\left(1+2^2+2^4\right)+...+2^{97}\left(1+2^2+2^4\right)\)

\(=21\left(2+2^7+...+2^{97}\right)⋮3\) và B chia hết cho 7

b: \(B=2\left(1+2^2\right)+2^5\left(1+2^2\right)+...+2^{97}\left(1+2^2\right)+2^{101}\)

Vì 2^101 chia 5 dư 2 nên B chia 5 dư 2