Chứng tỏ: A chia hết cho 13
A=3+3^2+3^3+...+3^99
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DN
5
N
6 tháng 10 2018
\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)....+\left(3^{97}+3^{98}+3^{99}\right)\)
\(A=3.\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)...+3^{97}.\left(1+3+3^2\right)\)
\(A=3.13+3^4.13+...+3^{97}.13\)
\(A=13.\left(3+3^4+..+3^{97}\right)⋮13\)
Vậy...
6 tháng 10 2018
\(A=3+3^2+3^3+...+3^{99}\)
\(A=\left(3+3^2+3^3\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\)
\(A=3\left(1+3+3^2\right)+...+3^{97}\left(1+3+3^2\right)\)
\(A=3\cdot13+...+3^{97}\cdot13\)
\(A=13\cdot\left(3+...+3^{97}\right)⋮13\left(đpcm\right)\)
HM
4
TT
20 tháng 10 2017
A=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^97+3^98+3^99)
A=3.(1+3+3^2)+3^4.(1+3+3^2)+...+3^97.(1+3+3^2)
A=3.13+3^4.13+...+3^97.13
A=13.(3+3^4+...+3^97) chia hết cho 13
20 tháng 10 2017
\(A=3+3^2+3^3+....+3^{99}\)
\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+.....+\left(3^{97}+3^{98}+3^{99}\right)\)
\(A=3.\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+...+3^{97}.\left(1+3+3^2\right)\)
\(A=3.13+3^4.13+....+3^{97}.13\)
\(A=13.\left(3+3^4+....+3^{97}\right)\)
\(\Leftrightarrow A⋮13\)
Vậy: \(A⋮13\)
Nhớ k cho mình nhé! Thank you!!!
7 tháng 10 2019
1)
a)\(\overline{ab}+\overline{ba}=10a+b+10b+a=11a+11b=11\left(a+b\right)⋮11\)
b) \(\overline{ab}-\overline{ba}=10a+b-10b-a=9a-9b=9\left(a-b\right)⋮9\)
2)
a) Có: \(\overline{abcd}=100\overline{ab}+\overline{cd}=99\overline{ab}+\left(\overline{ab}+\overline{cd}\right)\)
Mà \(\left\{{}\begin{matrix}99\overline{ab}⋮99\\\left(\overline{ab}+\overline{cd}\right)⋮99\end{matrix}\right.\)
\(\Rightarrow\overline{abcd}⋮99\)
b) Có: \(\overline{abcdef}=1000\overline{abc}+\overline{def}=999\overline{abc}+\left(\overline{abc}+\overline{def}\right)=37\cdot27\cdot\overline{abc}+\left(\overline{abc}+\overline{def}\right)\)
Mà \(\left\{{}\begin{matrix}37\cdot27\cdot\overline{abc}⋮37\\\left(\overline{abc}+\overline{def}\right)⋮37\end{matrix}\right.\)
\(\Rightarrow\overline{abcdef}⋮37\)
3)
a) Có: \(A=1+3+3^2+...+3^{1998}+3^{1999}+3^{2000}\\ A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{1998}+3^{1999}+3^{2000}\right)\\ A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1998}\left(1+3+3^2\right)\\ A=13+3^3\cdot13+...+3^{1998}\cdot13\\ A=13\left(1+3^3+...+3^{1998}\right)⋮13\)
b) Có: \(B=1+4+4^2+...+4^{2010}+4^{2011}+4^{2012}\\ B=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{2010}+4^{2011}+4^{2012}\right)\\ B=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{2010}\left(1+4+4^2\right)\\ B=21+4^3\cdot21+...+4^{2010}\cdot21\\ B=21\left(1+4^3+...+4^{2010}\right)⋮21\)
23 tháng 10 2015
TA CÓ:
A=30+3+32+33+........+311
(30+3+32+33)+....+(38+39+310+311)
3(0+1+3+32)+......+38(0+1+3+32)
3.13+....+38.13 cHIA HẾT CHO 13 NÊN A CHIA HẾT CHO 13( đpcm)
SG
0
10 tháng 11 2019
a)Ta có:A=3+32+33+...+318
=(3+32)+(33+34)+...+(317+318)
=3(1+3)+33(1+3)+...+317(1+3)
=3.4+33.4+...+317.4
Vì 4\(⋮\)4 nên 3.4+33.4+...+317.4\(⋮\)4
hay A\(⋮\)4
Ta có:A=3+32+33+...+318
=(3+32+33)+(34+35+36)+...+(316+317+318)
=3(1+3+32)+34(1+3+32)+...+316(1+3+32)
=3.13+34.13+...+316.13
Vì 13\(⋮\)13 nên 3.13+34.13+...+316.13\(⋮\)13
hay A\(⋮\)13
Vậy A chia hết cho 4, 13.
10 tháng 11 2019
A=3+32+33+...+318
A=(3+32)+(33+34)+...+(317+318)
A=3(1+3)+33(1+3)+...+317(1+3)
A=3x4+33x4+...+317x4
A=4x(1+33+...+317) chia hết cho 4
\(A=3+3^2+3^3+...+3^{99}\)
\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\)
`A =` \(\left(3+3^2+3^3\right).\left(1+3^3+...+3^{96}\right)\)
`A =` \(39.\left(1+3^3+...+3^{96}\right)\)
Mà `39 ⋮ 13`
`=> A ⋮ 13` (đpcm)