Ta có: \(1999^{30}=\left(1999^2\right)^5=8^{15}=\left(8^3\right)^5=16^5=1\) (mode[3][1])
\(\Rightarrow\left(1999^{30}\right)^{66}=1\) (mode[3][1])
\(\Rightarrow1999^{1980}=1\) (mode [3][1]) (1)
Ta lại có: \(1999^{21}=\left(1999^2\right)^{10}.1999=8^{10}.5=\left(8^5\right)^2.15=15\) (mode[3][1]) (2)
Từ (!) và (2) \(\Rightarrow1999^{1980}.1999^{21}=15\)
\(\Rightarrow1999^{2001}=15\) (mode[3][1])
hay \(1999^{2001}\) chia cho 31 dư 15