Đặt \(A=1+2+2^2+\cdots+2^{2005}\)

=>\(2A=2+2^2+2^3+\cdots+2^{2006}\)

=>\(2A-A=2+2^2+2^3+\ldots+2^{2006}-1-2-\cdots-2^{2005}\)

=>\(A=2^{2006}-1\)

Ta có: \(A=1+2+2^2+\cdots+2^{2005}\)

\(=1+2+\left(2^2+2^3+2^4\right)+\left(2^5+2^6+2^7\right)+\cdots+\left(2^{2003}+2^{2004}+2^{2005}\right)\)

\(=3+2^2\left(1+2+2^2\right)+2^5\left(1+2+2^2\right)+\cdots+2^{2003}\left(1+2+2^2\right)\)

\(=3+7\left(2^2+2^5+\cdots+2^{2003}\right)\)

=>A chia 7 dư 3

=>\(2^{2006}-1\) chia 7 dư 3

=>\(2^{2006}-1+1\) chia 7 dư 4

=>\(2^{2006}\) chia 7 dư 4

a)\(A=1+2+2^2+2^3+2^4+2^5+...+2^{2004}+2^{2005}+2^{2006}\)

\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{2004}+2^{2005}+2^{2006}\right)\)

\(A=7+2^3\left(1+2+2^2\right)+...+2^{2004}\left(1+2+2^2\right)\)

\(A=7+2^3.7+...+2^{2004}.7\)

\(A=7\left(1+2^3+...+2^{2004}\right)\) chia hết cho 7

b)\(2^{2006}=2^{2004}.2^2=\left(2^6\right)^{334}.4=64^{334}.4\)

Mặt khác: \(64\equiv1\left(mod7\right)\Rightarrow64^{334}\equiv1\left(mod7\right)\Rightarrow64^{334}.4\equiv4\left(mod7\right)\)

=>2²⁰⁰⁶ chia 7 dư 4

Trl :

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có lời giải ko bạn

Đặt S=1+2+2^2+..........+2^2019

Vì: S có 2020 số hạng nên ta chia S thành:673 nhóm mỗi nhóm có  3 số hạng và thừa 1 số hạng như sau 

S=1+(2+2^2+2^3)+(2^4+2^5+2^6)+...........+(2^2017+2^2018+2^2019)

S=1+2(1+2+4)+2^4(1+2+4)+........+2^2017(1+2+4)

S=1+2.7+2^4.7+.....+2^2017.7

S=1+7(2+2^4+2^2017) chia 7 dư 1

Vậy: 1+2+2^2+2^3+..........+2^2019 chia 7 dư 1

\(2^3\equiv1\left(mod7\right)\)

\(\Rightarrow\left(2^3\right)^{668}.2^2\equiv1^{668}.2^2\left(mod7\right)\)

\(\Rightarrow2^{2006}\equiv4\left(mod7\right)\)

-Vậy: \(2^{2006}\) chia 7 dư 4

\(2^{2006}=\left(2^{17}\right)^{118}=131072^{118}\)

Ma \(131072\equiv4\left(mod7\right)\)=>\(131072^{118}=4\left(mod7\right)\)

=> 131072^118 hay 2^2006 chia 7 du 3