Question

Tìm số dư trong phép chia số 2 mũ 2006 cho 7

Answer 1

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