cho M=1+20102+20103+20104+20105+20106+20107
Chung minh:M chia het cho 2011
Cho M = 1+2010+20102+20103+20104+20105+20106+20107
Chung minh:M chia het cho 2011
M=(2010+2010^2)+(2010^3+2010^4)+(2010^5+2010^6)+2010^7+1
=2010x2011+2010^3x2011+2010^5x2011+2010^7+1
=2011x(2010+2010^3+2010^5)+2010^7+1
mà 2010^6 đồng dư với 1 (mod 2011) nen 2010^6 x 2010 dong du voi 2010(mod 2011)
nên 2010^6 x 2010 +1 đồng dư với 2011 (mod 2011) nên 2010^7 +1 chia hết cho 2011 vậy m chia hết cho 2011
ai ủng hộ vài li-ke để lên hạng 3 đi ( tui sẽ trả li-ke lại )
m=1+2011+20112+20113+.....20117chung minh m chia het cho 2012
2013+2012^2(1+2012)+.......................+2011^6(1+2012) TA THẤY MOI SO DAU CO THUA SO 2012 +1 =2013 VAY NÓ CHIA HET CHO 13
1+2011=2012
VẦY TA CÓ 2011+1 + 2011^2+2011^2 X2011 +.......................2011^6 +2011^6 X 2011 SUUY RA 2012+2011^2(1+2011)+..........................+2016^6(1+2011)=(2011+1) X ( 2011^2+...............+2016^6) =2012(2011^2+...............+2016^6) TA THẤY 2012 CHIA HẾT CHO 2012 VẬY TỔNG NÀY CHIA HẾT CHO 2012
Cmr 10^2010-1 chia het cho 99
3^1930+2^1930 chia het cho 13
(2^10+1)^2010 chia het cho 25^2010
(30^4)^1975×15^1870×4^935-(7^5)^1954. Chia hết cho 23
12^2000-2^1000 chia hết cho 10
2011^2013+2013^2011 chia het cho 2012
Cho M=1+2010+2010 mũ 2+...+2010 mũ 7.Chung minh rang M chia het cho 2011
CMR:a)23!+19!-15! chia het cho 110
b)(10^28+8) chia het cho72
c)1+3+3^2+3^2011 chia het cho 10
chung minh rang: 1+3+3^2+3^3+...+3^2011 chia het cho 10
Ta có
1+3+32+33+...+32011
= (1+3+32+33)+....+(32008+32009+32010+32011)
=40+40+...+40
=10(4+4+...+4)\(⋮\)10 (đpcm)
đặt A= 1+3+32 +........+32011
=> 3A=3+32 +33+.......+32011+32012
=> 3A-A=32012-1
=>A=(32012-1)/2
Đặt \(A=1+3+3^2+3^3+.......+3^{2011}\)
\(\Rightarrow A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+........+\left(3^{2008}+3^{2009}+3^{2010}+3^{2011}\right)\)
\(\Rightarrow A=10+3^4.\left(1+3+3^2+3^3\right)+.......+3^{2008}.\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=10+3^4.10+.........+3^{2008}.10\)
\(\Rightarrow A=10\left(1+3^4+......+3^{2008}\right)⋮10\)( đpcm )
Vậy .....
cho S= 3^1+3^3+3^5+....+3^2011+3^2013+3^2015
chứng minh rằng S ko chia het cho9 va chia het cho 70
giai ro rang ra giup mik nha
Xet tich gom 11 thua so : A = (5a.2006.b)(6a.2005.b)(7a.2007.b)....(15a.1996.b) voi a>b ; a,b la cac so tu nhien. CMR neu A chia het cho 2011 thi A chia het cho 2011^11
cho S=7^2013-7^2012+7^2011-7^2010+...-7^2+7-1.cm S chia het cho 6
S = 72013 - 72012 + 72011 - 72010 + ....+ 73 - 72 + 7 - 1
= ( 72013 - 72012 ) + ( 72011 - 72010 ) + ....+ ( 73 - 72 ) + ( 7 - 1 )
= 72012 ( 7 - 1 ) + 72010 ( 7 - 1 ) + .... + 72 ( 7 - 1 ) + ( 7 - 1 )
= 72012.6 + 72010.6 + .... + 72.6 + 6
= 6.( 72012 + 72010 + .... + 72 + 1 ) chia hết cho 6 ( đpcm )