Question

tìm số dư A=3^0+3^1+3^2+...+3^2016 cho 52

Answer 1

\(A=3^0+3^1+3^2+...+3^{2016}\\ =1+3^1+3^2+...+3^{2016}\)

Gọi \(3^1+3^2+3^3+...+3^{2016}\) là \(D\)

\(D=3^1+3^2+3^3+...+3^{2016}\\ =\left(3^1+3^2+3^3+3^4+3^5+3^6\right)+\left(3^7+3^8+3^9+3^{10}+3^{11}+3^{12}\right)...+\left(3^{2011}+3^{2012}+3^{2013}+3^{2014}+3^{2015}+3^{2016}\right)\\ =3^1\cdot\left(1+3+3^2+3^3+3^4+3^5\right)+3^7\cdot\left(1+3+3^2+3^3+3^4+3^5\right)+...+3^{2011}\cdot\left(1+3+3^2+3^3+3^4+3^5\right)\\ =3^1\cdot364+3^7\cdot364+...+3^{2011}\cdot364\\ =364\cdot\left(3^1+3^7+...+3^{2011}\right)\\ =52\cdot7\cdot\left(3^1+3^7+...+3^{2011}\right)⋮52\)\(A=1+D=1+52k\left(k⋮N\right)\) chia 52 dư 1

Vậy A chia 52 dư 1