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Ta có:
2008100+200899
= 200899.2008+200899
= 200899.(2008+1)
= 200899.2009
Vì 2009 chia hết cho 2009 => 200899.2009 chia hết cho 2009 => 2008100+200899 chia hết cho 2009
2008100 + 200899 = 200899(2008+1) = 200899. 2009
=> 2008100 + 200899 chia hết cho 2009
\(2008^{99}\cdot2008+2008^{99}\)
\(=2008^{99}\cdot\left(2008+1\right)\)
\(=2008^{99}.2009⋮2009\left(dpcm\right)\)
Ta có: \(A=1\cdot2\cdot3\cdot...\cdot2007\cdot2008\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
\(A=2008!\left[\left(1+\frac{1}{2008}\right)+\left(\frac{1}{2}+\frac{1}{2007}\right)+...+\left(\frac{1}{1004}+\frac{1}{1005}\right)\right]\)
\(A=2008!\left(\frac{2009}{2008}+\frac{2009}{2\cdot2007}+...+\frac{2009}{1004\cdot1005}\right)\)
\(A=\frac{2009!}{2008}+\frac{2009!}{2\cdot2007}+...+\frac{2009!}{1004\cdot1005}\)
\(A=2009\left(2\cdot3\cdot...\cdot2017+3\cdot4\cdot...\cdot2016\cdot2018+2\cdot3\cdot...\cdot1003\cdot1006\cdot...\cdot2018\right)\)
chia hết cho 2019
=> đpcm
Dễ quá, thực hiện qui tắc bỏ dấu ngoặc được:
\(2009+2009^2+....+2009^{2009}-1-2009-...-2009^{2008}\)
\(=-1+\left(2009-2009\right)+\left(2009^2-2009^2\right)+...+\left(2009^{2008}-2009^{2008}\right)+2009^{2008}\)
\(=2009^{2008}-1\)
\(=\left(2009-1\right)\left(2009^{2007}+2009^{2008}+...+2009+1\right)\)
\(=2008\left(2009^{2007}+2009^{2008}+...+2009+1\right)\) chia hết cho 2008
=> ĐPCM
Chứng Minh Rằng: (2009+20092+20093+20094+...+20092009)-(1+2009+20092+20093+...+20092008) chia hết cho 2008.
Đặt A=2009+20092+20093+20094+...+20092009, B=1+2009+20092+20093+20094+...+20092008
Ta có:
+)A=2009+20092+20093+20094+...+20092009
2009A= 20092+20093+20094+...+20092010
2009A-A=(20092+20093+20094+...+20092010)-(2009+20092+20093+20094+...+20092009)
2008A=20092010- 2009
=> A=(20092010- 2009)/2008
=> A chia hết cho 2008.
B=1+2009+20092+20093+20094+...+20092008
2009B=2009+20092+20093+20094+...+20092010
2009B-B=(2009+20092+20093+20094+...+20092010)-(1+2009+20092+20093+20094+...+20092009)
2008B=20092010-1
=>B=(20092010-1)/2008
=>B chia hết cho 2008
=> A-B chia hết cho 2008.
=> ĐPCM
max dễ :
10 chia 3 dư 1 , suy ra 10^n chia 3 dư 1^n
suy ra 10^n chia 3 dư 1
ta có : 4 chia 3 dư 1
suy ra 10^n-4 chia 3 dư 1-1
10^n-4 chia 3 dư 0
10^n-4 chia het cho 3