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102017 + 102016 + 102015
= 102015 . ( 102 + 10 + 1 )
= 102015 . 111
= 102014 . 10 . 111
= 102014 . 5 . 2 . 111
= 102014 . 2 . 555 \(⋮555\)
Vậy 102017+ 22016 + 102015 \(⋮555\)
Ta có\(\hept{\begin{cases}10^{2017}⋮5\\10^{2016}⋮5\\10^{2015}⋮5\end{cases}}\)
\(\Rightarrow10^{2017}+10^{2016}+10^{2015}⋮5\left(1\right)\)
Lại có\(10^{2017}+10^{2016}+10^{2015}=10^{2015}\left(100+10+1\right)=10^{2015}.111⋮111\left(2\right)\)
Mặt khác\(\left(5,111\right)=1\left(3\right)\)
Từ\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow10^{2017}+10^{2016}+10^{2015}⋮555\left(đpcm\right)\)
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A = 2016 + 2016 2 + 2016 3 + 2016 4 + 2016 5 + 2016 6 + 2016 7 + 2016 8 + 2016 9 + 2016 10
A = 2016 . 1 + 2016 . 2016 + 20163 . 1 + 20163 . 2016 + 20165 . 1 + 20165 . 2016 + 20167 . 1 + 20167 . 2016 + 20169 . 1 + 20169 . 2016
A = 2016 . ( 1 + 2016 ) + 20163 . ( 1 + 2016 ) + 20165 . ( 1 + 2016 ) + 20167 . ( 1 + 2016 ) + 20169 . ( 1 + 2016 )
A = 2016 . 2017 + 20163 . 2017 + 20165 . 2017 + 20167 . 2017 + 20169 . 2017
A = ( 2016 + 20163 + 20165 + 20167 + 20169 ) . 2017 chia hết cho 2017
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3:
a) Ta có:
44433 = 4443 . 11 = (4443)11 = 87 528 38411
33344 = 3334 . 11 = (3334)11 = 12 296 370 32111
Vì 87 528 38411 < 12 296 370 32111 nên 44433 < 33344
Vậy,...
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a) \(2^{2017}+2^{2014}=2^{2014}\left(2^3+1\right)=2^{2014}.9⋮9\)
b) \(4^{2016}+4^{2014}=4^{2014}\left(4^2+1\right)=4^{2014}.17\)
2) \(3.4^{n+2}+4^n=49\\ \Rightarrow4^n\left(3.4^2+1\right)=49\\ \Rightarrow4^n.33=49\\ \Rightarrow4^n=16\\ \Rightarrow n=2\)
3) \(200-180:\left[36.5-7.25\right]\\ =200-180:\left[180-175\right]\\ =200-180:5\\ =200-36\\ =164\)
(102 . 132016 + 69.132016) : 132017
= 132016.( 100 + 69) : 132017
= 132016. 169 : 132017
= 169 : 13
= 13
102 . 132016 + 69.132016) : 132017
= 132016.( 100 + 69) : 132017
= 132016. 169 : 132017
= 169 : 13
= 13