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a)2004100+200499=200499(2004+1)=201499.2005
=>201499.2005chia hết cho 2005
=> 2004100+200499 chia hết cho 2005
b) 413+325-88
=(22)13+(25)5-(23)8
=226+225-224
=224(22+2-1)
=225.5
=>225chia hết cho 5 => 413+325-88 chia hết cho 5
a) \(2010^{100}+2010^{99}\)
\(=2010^{99}\left(2010+1\right)\)
\(=2010^{99}.2011⋮2011\left(dpcm\right)\)
b) \(3^{1994}+3^{1993}-3^{1992}\)
\(=3^{1992}\left(3^2+3-1\right)\)
\(=3^{1992}.11⋮11\left(dpcm\right)\)
c) \(4^{13}+32^5-8^8\)
\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)
\(=2^{26}+2^{25}-2^{24}\)
\(=2^{24}\left(2^2+2-1\right)\)
\(=2^{24}.5⋮5\left(dpcm\right)\)
a)2004100+200499=200499(2004+1)=201499.2005
=>201499.2005chia hết cho 2005
=> 2004100+200499 chia hết cho 2005
b) 413+325-88
=(22)13+(25)5-(23)8
=226+225-224
=224(22+2-1)
=225.5
=>225chia hết cho 5
=> 413+325-88 chia hết cho 5
a) Ta có A = 710 + 79 - 78
= 78( 72 + 7 - 1 )
= 78 . 55 ⋮ 11 vì 55 ⋮ 11
Vậy A ⋮ 11
b) Ta có B = 115 + 114 + 113
= 113( 112 + 11 + 1 )
= 113 . 133 ⋮ 7
Vậy B ⋮ 7
a,A=710+79-78=78(72+7-1)=78x55 ⋮11 vì 55⋮11
b,115+114+113=113(112+11+1)=113x133⋮7 vì 133⋮7
\(A=8+8^3+8^5+....+8^{99}+2017.\)
\(=\left(8+8^3\right)+\left(8^5+8^7\right)+....+\left(8^{98}+8^{99}\right)+2017\)
\(=8\left(1+8^2\right)+8^5\left(1+8^2\right)+.....+8^{98}\left(1+8^2\right)+2017\)
\(=8.65+8^5.65+....+8^{98}.65+2017\)
\(=65\left(8+8^5+....+8^{98}\right)+2017\)
\(=13.5\left(8+8^5+...+8^{98}\right)+2017\)
\(\hept{\begin{cases}13.5\left(8+8^5+...+8^{98}\right)⋮5\\2017⋮̸⋮5\end{cases}\Rightarrow}đpcm\)