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a) \(1991\equiv2\left(mod9\right)\)
=> \(1991^{1990}\equiv2^{1990}\left(mod9\right)\)
=> \(1991^{1990}\equiv2^{3.633}.2\left(mod9\right)\equiv-2\left(mod9\right)\)
\(1990^{1991}\equiv1\left(mod9\right)\)
=> \(1991^{1990}+1990^{1991}\equiv8\left(mod9\right)\)
=> đpcm
b) Ta có 89 là số lẻ =>8926 lẻ
45 là số lẻ => 4521lẻ
=> 8926 - 4521 chẵn => chia hết cho 2 => đpcm
NHỚ CHO MIK NHA BẠN THÂN MẾN
mod là modun
ví dụ như 3 chia 2 dư 1
5 chia 2 dư 1 ta nói 3 đồng dư với 1 theo modun 2
và \(5\equiv1\left(mod2\right)\)
a) 56 - 5 + 54
= 54 . (52 - 5 + 1)
= 54 . (25 - 5 + 1)
= 54 . 21
= 54 . 3 . 7 chia hết cho 7
=> đpcm
b) 121996 - 21000
= (124)499 - (24)250
= (...6)499 - (...6)250
= (...6) - (...6)
= (...0) chia hết cho 10
=> đpcm
Ủng hộ mk nha ♡_♡☆_☆
a) \(5^6-5^5+5^4\)
\(=5^4\left(5^6-5+1\right)\)
\(=5^4\left(25-5+1\right)\)
\(=5^4.21\)
\(=7.3.5^4\)chia hết cho 7
b) \(12^{1996}-2^{1000}\)
\(=\left(12^4\right)^{499}-\left(2^4\right)^{250}\)
\(=\left(..6\right)^{499}-\left(...6\right)^{250}\)
\(=\left(....0\right)\)chia hết cho 10
Ủng hộ nha
Theo Fermat:a^11=a(mod 11)=>a^1991=a(mod 11)
tick nha
a,Ta có : \(1996\equiv1\left(mod5\right)\)
\(\Rightarrow1996^{1996}\equiv1^{1996}\left(mod5\right)\)
\(1991\equiv1\left(mod5\right)\)
\(\Rightarrow1991^{1991}\equiv1^{1991}\left(mod5\right)\)
\(\Rightarrow1996^{1996}-1991^{1991}\equiv1^{1996}-1^{1991}\left(mod5\right)\)
\(\Leftrightarrow1996^{1996}-1991^{1991}\equiv0\left(mod5\right)\)
Hay \(1996^{1996}-1991^{1991}⋮5\)
b,Ta có : \(9^{1972}=\left(9^2\right)^{986}=81^{986}\)
\(7^{1972}=\left(7^4\right)^{493}=2401^{493}\)
Ta lại có : \(81\equiv1\left(mod10\right)\)
\(\Rightarrow81^{986}\equiv1^{986}\left(mod10\right)\)
\(2401\equiv1\left(mod10\right)\)
\(\Rightarrow2401^{493}\equiv1^{493}\left(mod10\right)\)
\(\Rightarrow9^{1972}-7^{1972}=81^{986}-2401^{493}\equiv1^{986}-1^{493}\left(mod10\right)\)
\(\Leftrightarrow9^{1972}-7^{1972}=81^{986}-2401^{493}\equiv0\left(mod10\right)\)
hay \(9^{1972}-7^{1972}⋮10.\)
c, Ta có : \(89\equiv1\left(mod2\right)\)
\(\Rightarrow89^{26}\equiv1^{26}\left(mod2\right)\)
\(45\equiv1\left(mod2\right)\)
\(\Rightarrow45^{21}\equiv1^{21}\left(mod2\right)\)
\(\Rightarrow89^{26}-45^{21}\equiv1^{26}-1^{21}\left(mod2\right)\)
\(\Rightarrow89^{26}-45^{21}\equiv0\left(mod2\right)\)
Hay \(89^{26}-45^{21}⋮0\)
\(1996\equiv1\left(mod5\right)\Rightarrow1996^{1996}\equiv1\left(mod5\right)\)
\(1991\equiv1\left(mod5\right)\Rightarrow1991^{1991}\equiv1\left(mod5\right)\)
\(\Rightarrow1996^{1996}-1991^{1991}\equiv1-1=0\left(mod5\right)\Leftrightarrowđpcm.\)
\(9^{1972}=\left(9^2\right)^{986}=81^{986}\equiv1\left(mod10\right)\)
\(7^{1972}=\left(7^4\right)^{493}=2401^{493}\equiv1\left(mod10\right)\)
\(\Rightarrowđpcm.\)