Ta có: \(19^2\equiv1\left(mod10\right)\)
\(\left(19^2\right)^{1002}\equiv1^{1002}\equiv1\left(mod10\right)\)
\(\Rightarrow19^{2004}\cdot19\equiv1\cdot9\equiv9\left(mod10\right)\) (*)
Ta có: \(11\equiv1\left(mod10\right)\)
\(11^{2004}\equiv1^{2004}\equiv1\left(mod10\right)\)(**)
Từ (*);(**)
=> \(A=19^{2005}+11^{2004}\equiv9+1\equiv10\left(mod10\right)\)
=> A⋮10(đpcm)
Ta có: \(19^{2015}=19^{2014}.19=\left(19^2\right)^{1007}.19=\left(...1\right)^{1007}.19=\left(...1\right).19=\left(...9\right)\)
Và \(11^{2014}=\left(...1\right)\)
\(\Rightarrow19^{2015}+11^{2014}=\left(...9\right)+\left(...1\right)=\left(...0\right)⋮10\)
\(\Rightarrow A\) \(⋮\) \(10\)
Vậy \(A\) \(⋮\) \(10.\)
