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a, Muon chung minh A\(⋮24\)thì A\(⋮8va3\)
+)Ta có A=\(\overline{11111...016⋮8}\)vi 3 chu so tc la \(016⋮8\)
+)Lại có A có tong cac chu so la 1+1+1+1+1+1+6=12\(⋮3\)nên A\(⋮3\)
Vay a\(⋮3x8=24\)
b,cac so cua A deu chia het cho 102 mà 16 ko chia het cho 102 nen A ko la so chih phuong
\(A=16+10^{2015}+10^{2016}+10^{2017}+10^{2018}\)
Ta có :
\(10^x\) có dạng \(3k+1\)
\(\Leftrightarrow10^{2015}+10^{2016}+10^{2017}+10^{2018}=\left(3k+1\right).4=12k+1\)
Mà \(16=3k+1\)
\(\Leftrightarrow A=\left(12k+1\right)+\left(3k+1\right)\)
\(\Leftrightarrow A=3k+1\)
\(\Leftrightarrow A\) ko là số chính phương
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\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow10^{2017}+10^{2016}+10^{2015}⋮555\left(đpcm\right)\)
\(a,19^{2018}+13^{2018}\)
\(19\equiv-1\left(mod10\right)\)
\(\Rightarrow19\equiv\left(-1\right)^{2018}=1\left(mod10\right)\)
\(13^{2018}=\left(13^2\right)^{1009}=169^{1009}\)
\(169\equiv-1\left(mod10\right)\)
\(\Rightarrow169^{1009}\equiv\left(-1\right)^{1009}=-1\left(mod10\right)\)
\(\Rightarrow19^{2018}+13^{2018}\equiv1+\left(-1\right)=0\left(mod10\right)\)
\(\Leftrightarrow19^{2018}+13^{2018}⋮10\left(đpcm\right).\)
\(b,17^{2013}+23^{2017}\)
\(17^{2013}=\left(17^2\right)^{1006}.17=289^{1006}.17\)
\(289\equiv-1\left(mod10\right)\)
\(\Rightarrow289^{1006}\equiv\left(-1\right)^{1006}=1\left(mod10\right)\)
\(17\equiv7\left(mod10\right)\)
\(\Rightarrow289^{1006}.17\equiv1.7=7\left(mod10\right)\)( 1 )
\(23^{2017}=\left(23^2\right)^{1008}.23=529^{1008}.23\)
\(529\equiv-1\left(mod10\right)\)
\(\Rightarrow529^{1008}\equiv\left(-1\right)^{2018}=1\left(mod10\right)\)
\(23\equiv3\left(mod10\right)\)
\(\Rightarrow529^{1008}.23\equiv1.3=3\left(mod10\right)\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow17^{2013}+23^{2017}\equiv7+3=10\left(mod10\right)\)
Mà \(10⋮10\Rightarrow17^{2013}+23^{2017}\equiv0\left(mod10\right)\)
\(\Leftrightarrow17^{2013}+23^{2017}⋮10\left(đpcm\right).\)
\(c,17^5+24^4-13^{21}\)
\(=\overline{...7}+\overline{...6}-\overline{...3}\)
\(=\overline{...0}⋮10\)
\(\Rightarrow17^5+24^4-13^{21}⋮10\left(đpcm\right).\)