Chứng minh rằng: 201.202.203.....600 chia hết cho 3^200
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Ta có:
\(742^3-692^3=\left(742-692\right)\left(742^2+742.692+692^2\right)=50.\left(742^2+742.692+692^2\right)\)
Do \(742⋮2\Rightarrow742^2⋮4\)
\(\left\{{}\begin{matrix}742⋮2\\692⋮2\end{matrix}\right.\) \(\Rightarrow742.692⋮4\)
\(692⋮2\Rightarrow692^2⋮4\)
\(\Rightarrow\left(742^2+742.692+692^2\right)⋮4\)
\(\Rightarrow\left(742^3-692^3\right)⋮\left(50.4=200\right)\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
1993 - 199 = 199 ( 1992 - 1 ) = 199 ( 199 + 1 ) ( 199 - 1 ) = 199 . 198 . 200
=> 1993 - 199 chia hết cho 200
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1993-199=199(1992-1)=199(199+1)(199-1)=199.198.200 chia hết cho 200
CHọn mình nha :)
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A= 1+2+3+...+1995
=1995+(1+1994)+(2+1993)+...+(996+999)+(997+998)
=1995+1995+1995+...+1995+1995
=1995x998\(⋮1995\)
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\(21^{10}-1\)
\(=\left(20+1\right)^{10}-1\)
\(=20^{10}+1^{10}-1\)
\(=20^{10}+\left(1-1\right)\)
\(=\left(20^2\right)^5\)
\(=400^5\)
\(=\left(200.2\right)^5\)
\(=200^5.2^5⋮200\left(đpcm\right)\)
21^10 -1
=(21^5)^2-1^2
=(21^5+1)(21^5-1)
Có 21^5+1=B suy rađặt 21^5+1=2k
suy ra 21^10=2k(21^5-1)=2k
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)