Gọi số tự nhiên đó là a, ta có:

\(a=11k+6\)\(\Rightarrow a+27=11k+6+27=11k+33\)\(=11\cdot\left(k+3\right)\)

\(a=4q+1\Rightarrow a+27=4q+1+27=4q+28\)

\(=4\cdot\left(q+7\right)\)

\(a=19m+11\Rightarrow a+27=19m+11+27=19m+38\)

\(=19\cdot\left(m+2\right)\)

Mà \(a+27⋮11;4;19\)

Mà a là số tự nhiên nhỏ nhất \(a+27\in BCNN\left(11;4;19\right)\)

BCNN(11;4;19)=836

\(\Rightarrow a+27=836\)

\(\Rightarrow a=836-27\)

\(\Rightarrow a=809\)

Vậy a=809

Ta thấy:\(\dfrac{232323}{494949}=\dfrac{23\cdot10101}{49\cdot10101}=\dfrac{23}{49}\)

Vậy \(\dfrac{232323}{494949}=\dfrac{23}{49}\)

Vì \(2017^{18}+1>2017^{17}+1\)

\(\Rightarrow B=\dfrac{2017^{18}+1}{2017^{17}+1}>1\)

\(\Rightarrow B=\dfrac{2017^{18}+1}{2017^{17}+1}< \dfrac{2017^{18}+1+2016}{2017^{17}+1+2016}=\dfrac{2017^{18}+2017}{2017^{17}+2017}=\dfrac{2017\cdot\left(2017^{17}+1\right)}{2017\cdot\left(2017^{16}+1\right)}=A\)

Vậy B<A hay A>B

Ta có: \(5+n^2-2n⋮n-2\)

\(\Rightarrow5+n\cdot n-2\cdot n⋮n-2\)

\(\Rightarrow5+n\cdot\left(n-2\right)⋮n-2\)

Vì \(5+n\cdot\left(n-2\right)⋮n-2\)

\(\Rightarrow n\cdot\left(n-2\right)⋮n-2\)

hay \(5⋮n-2\)

\(\Rightarrow5\inƯC\left(n-2\right)\) hoặc \(5\in\left\{-5;-1;1;5\right\}\)

Ta có bảng sau:

Vậy \(n\in\left\{-3;1;3;7\right\}\)