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![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)
\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)
\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)
\(S=4\left(3^2+3^4+3^6+3^8\right)\)
\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\\=(3+3^2)+(3^3+3^4)+(3^5+3^6)+(3^7+3^8)\\=3\cdot(1+3)+3^3\cdot(1+3)+3^5\cdot(1+3)+3^7\cdot(1+3)\\=3\cdot4+3^3\cdot4+3^5\cdot4+3^7\cdot4\\=4\cdot(3+3^3+3^5+3^7)\)
Vì \(4\cdot(3+3^3+3^5+3^7) \vdots 4\)
nên \(B\vdots4\).
`#3107.101107`
\(B=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+\left(3^7+3^8\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+3^5\left(1+3\right)+3^7\left(1+3\right)\)
\(=\left(1+3\right)\left(3+3^3+3^5+3^7\right)\)
\(=4\left(3+3^3+3^5+3^7\right)\)
Vì \(4\left(3^3+3^5+3^7\right)\) $\vdots 4$
`\Rightarrow B \vdots 4`
Vậy, `B \vdots 4.`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 2^35.(1+2+2^2+2^3) = 2^35.15 = 2^35.5.3 chia hết cho 3
=> A chia hết cho 3
k mk nha
\(A=2^{35}+2^{36}+2^{37}+2^{38}\)
\(A=2^{35}+2^{35}.2+2^{37}+2^{37}.2\)
\(A=2^{35}\left(1+2\right)+2^{37}\left(1+2\right)\)
\(A=2^{35}.3+2^{37}.3\)
\(A=3\left(2^{35}+2^{37}\right)\)CHIA HẾT CHO 3
![](https://rs.olm.vn/images/avt/0.png?1311)
Tó biết làm mỗi 2 bài trên thui
1 ) aaa aaa = a . 111 111 = a . 11 . 10101 => chia hết cho 11
2 ) abc abc = abc . 1001 = abc . 11 . 91 = > chia hết cho 11
làm theo cách thầy dạy chứ hoàn toàn ko nhìn sách giải nhé
bạn cứ nhóm cặp phù hợp rồi làm thôi
là sao hả bạn?