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a) \(\overline{aaaaaa}=a.111111=a.3.37037\) \(⋮\)\(37037\)
b) Nhận thấy các hạng tử trong B đều chia hết cho 3 => B chia hết cho 3
\(B=3+3^3+3^5+3^7+...+3^{2017}+3^{2019}+3^{2021}\)
\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+....+\left(3^{2017}+3^{2019}+3^{2021}\right)\)
\(=3\left(1+3^2+3^4\right)+3^7\left(1+3^2+3^4\right)+...+3^{2017}\left(1+3^2+3^4\right)\)
\(=\left(1+3^2+3^4\right)\left(3+3^7+...+3^{2017}\right)\)
\(=91\left(3+3^7+....+3^{2017}\right)\)\(⋮\)\(91\)
mà (3;91) = 1
=> B chia hết cho 273
B chia hết cho 273
Còn câu a thì mình không biết nhé, xin lỗi bạn.
a) \(A=5+5^2+5^3+...+5^8\)
\(=\left(5+5^2\right)+5^2\cdot\left(5+5^2\right)+...+5^6\cdot\left(5+5^2\right)\)
\(=\left(5+5^2\right)\cdot\left(1+5^2+...+5^6\right)\)
\(=30\cdot\left(1+5^2+...+5^6\right)\)chia hết cho 30.
b) \(B=3+3^3+3^5+3^7+...+3^{29}\)
\(=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{26}\cdot\left(3+3^3+3^5\right)\)
\(=\left(3+3^3+3^5\right)\cdot\left(1+3^6+...+3^{26}\right)\)
\(=273\cdot\left(1+3^6+3^{26}\right)\)chia hết cho 273.
\(A=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{25}+3^{27}+3^{29}\right)\)
\(=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^3+3^5\right)\)
\(=273+3^6.273+........+3^{24}.273\)
\(=273\left(1+3^6+......+3^{24}\right)\)chia hết cho 273
Ta có:
273=3+3^3+3^5
A=(3+3^3+3^5)+(3^7+3^9+3^11)+...+(3^25+3^27+3^29)
A=1×(3+3^3+3^5)+3^6×(3+3^3+3^5)+...+3^24×(3+3^3+3^5)
A=1×273+3^6×273+...+3^24×273
A=(1+3^6+...+3^24)×273
Suy ra: A chia hết cho 273
\(B=3+3^3+3^5+3^7+3^9+...+3^{39}\)
\(B=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{34}\left(3+3^3+3^5\right)\)
\(B=\left(3+3^3+3^5\right)\left(1+3^6+...+3^{34}\right)\)