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CMR( 17^8+24^4+13^21) chia hết cho 10
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a) Ta có : 51n=\(\overline{...1}\)
47102=472.(474)25=\(\left(\overline{...9}\right).\left(\overline{...1}\right)=\overline{...9}\)
\(\Rightarrow51^n+47^{102}=\left(\overline{...1}\right)+\left(\overline{...9}\right)=\overline{...0}⋮10\)
Vậy 51n+47102\(⋮\)10.
b) Ta có : \(17^5=17.17^4=17.\left(\overline{...1}\right)=\overline{...7}\)
\(24^4=\overline{...6}\)
\(13^{21}=13.\left(13^4\right)^5=13.\left(\overline{...1}\right)=\overline{...3}\)
\(\Rightarrow17^5+24^4-13^{21}=\left(\overline{...7}\right)+\left(\overline{...6}\right)-\left(\overline{...3}\right)=\overline{...0}⋮10\)
Vậy 175+244+1321\(⋮\)10
Ta có:
175 + 244 -1321
= 17 . 174 + ...6 - 134.5 . 13
= ...7 + ...6 - ...3
= ...0
Vì 175 + 244 - 1321 có chữ số tận cùng là 0 nên 175 + 244 + 1321 cũng chia hết cho 10.
175+244-1321=(174).17+(.........6)-(134)5.13=(................1).17+(................6)-(...............1).13
=(..........7)+(........6)-(..............1)=......................0 chia hết cho 10
=>175+244-1321 chia hết cho 10
= ( ...1 ) + (19992)999 · 1999
= (...1) + (...1)999 · 1999
= (...1 ) + (...1) · 1999
= (...1 ) + (...1999)
=(...2000 ) \(⋮10\)
b) 162001 - 82000
= ( ... 6 )2001 - ( 82 )1000
= (...6) - (...6)1000
=(...6 ) - (...6 )
= ( ...0 ) \(⋮10\)
c) 192005 + 112004
= ( 192 )1002 · 19 + (...1 )2004
= ( ... 1 )1002 · 19 + (...1)
= ( ...1 ) · 19 + (...1 )
= (...19 ) + (...1)
=( ... 20 ) \(⋮10\)
d) 175 + 244 - 1321
= (...74) · 17 + (..42 )2 - ( ...34 )5 · 13
= (...1) · 17 + ( ...6 )2 - ( ...1)5 · 13
= ( ...17 ) + (...6 ) - (...1) · 13
= (...23 ) - (..13)
= (...10 )\(⋮10\)
câu a) https://olm.vn/hoi-dap/question/228404.html
câu b)hhttps://olm.vn/hoi-dap/question/228404.html
c)https://olm.vn/hoi-dap/question/721614.html
d)https://olm.vn/hoi-dap/question/721614.html
\(51^n+47^{102}\)
\(=\overline{.....1}+\overline{.....9}\)
\(=\overline{.....0}⋮10\)
\(17^5+24^4-13^{21}\)
\(=\overline{....7}+\overline{...6}-\overline{.....3}\)
\(=\overline{.....0}⋮10\)