\(2^{2019}=2^4.2^{2015}=16.2^{2015}=16.\left(2^5\right)^{403}=16.\left(32\right)^{403}\)
Do \(32\equiv-1\left(mod11\right)\Rightarrow32^{403}\equiv\left(-1\right)^{403}\left(mod11\right)\)
\(\Rightarrow32^{403}\equiv-1\left(mod11\right)\)
\(\Rightarrow16.32^{403}\equiv-16\left(mod11\right)\)
\(-16=-22+5\Rightarrow-16\) chia 11 dư 5
Vậy \(2^{2019}\) chia 11 dư 5