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a) chứng tỏ : abcabc chia hết cho 11
Ta có 123123:11=11193
Vậy abcabc chia hết cho 11
b)\(\frac{9\cdot15\cdot21\cdot12\cdot20}{5\cdot6\cdot45\cdot18\cdot4}=\frac{9\cdot3\cdot5\cdot3\cdot7\cdot2\cdot2\cdot3\cdot2\cdot2\cdot5}{5\cdot2\cdot3\cdot5\cdot3\cdot3\cdot2\cdot3\cdot3\cdot2\cdot2}\)\(=\frac{7\cdot5}{3}=\frac{35}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,A=7^{15}+7^{16}+7^{17}\)
\(A=7^{15}\left(1+7+7^2\right)\)
\(A=7^{15}.57\)
Ta có :
\(A=7^{15}.57⋮57\)
\(\Rightarrow A⋮57\)
\(b,B=2+2^2+2^3+....+2^{60}\)
\(B=\left(2+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(B=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(B=2.7+...+2^{58}.7\)
\(B=7\left(2+2^4+....+2^{58}\right)\)
Ta có :
\(B=7\left(2+2^4+....+2^{58}\right)⋮7\)
\(\Rightarrow B⋮7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:vì 15 chia hết cho 5 suy ra 2022.15 chia hết cho 5
vì 25 chia hết cho 5 suy ra 2022.15 + 25 chia hết cho 5
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1) \(1+4+4^2+4^3+...+4^{2012}\)
\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{2010}+4^{2011}+4^{2012}\right)\)
\(=21+21\cdot4^3+...+21\cdot4^{2010}\)
\(=21\cdot\left(1+4^3+...+4^{2010}\right)\) chia hết cho 21
2) \(1+7+7^2+7^3+...+7^{101}\)
\(=\left(1+7\right)+\left(7^2+7^3\right)+...+\left(7^{100}+7^{101}\right)\)
\(=8+8\cdot7^2+...8\cdot7^{100}\)
\(=8\cdot\left(1+7^2+...+7^{100}\right)\) chia hết cho 8
3) CM chia hết cho 5:
\(2+2^2+2^3+2^4+...+2^{100}\)
\(=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{98}+2^{100}\right)\)
\(=5\cdot2+5\cdot2^2+...+5\cdot2^{98}\)
\(=5\cdot\left(2+2^2+...+2^{98}\right)\) chia hết cho 5
CM chia hết cho 31:
\(2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\cdot31+...+2^{96}\cdot31\)
\(=31\cdot\left(2+...+2^{96}\right)\) chia hết cho 31