Question

A=1+2+2²+2³+...+2¹¹⁹ chia hết cho 3,7,17,31

Answer 1

\(A=1+2+2^2+2^3+...+2^{119}\)

\(\Rightarrow A=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{118}+2^{119}\right)\)

\(\Rightarrow A=\left(1+2\right)+2^2\left(1+2\right)+...+2^{118}\left(1+2\right)\)

\(\Rightarrow A=\left(1+2\right)\left(1+2^2+...+2^{118}\right)\)

\(\Rightarrow A=3\left(1+2^2+...+2^{118}\right)⋮3\)

Answer 2

\(A=\left(1+2\right)+2^2\left(1+2\right)+...+2^{118}\left(1+2\right)\)

\(=3\left(1+...+2^{118}\right)⋮3\)

\(A=\left(1+2+2^2\right)+...+2^{117}\left(1+2+2^2\right)\)

\(=7\left(1+...+2^{117}\right)⋮7\)

\(A=\left(1+2+2^2+2^3+2^4\right)+...+2^{115}\left(1+2+2^2+2^3+2^4\right)\)

\(=31\left(1+...+2^{115}\right)⋮31\)

Answer 3

\(A=1+2+2^2+2^3+...+2^{119}\)

\(\Rightarrow A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{117}+2^{118}+2^{119}\right)\)

\(\Rightarrow A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{117}\left(1+2+2^2\right)\)

\(\Rightarrow A=\left(1+2+2^2\right)\left(1+2^3+...+2^{117}\right)\)

\(\Rightarrow A=17\left(1+2^3+...+2^{117}\right)⋮17\)

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