Tìm số dư trong phép chia: \(2^{1999}\) cho 35
Ta có: \(2^{24}\equiv1\left(mod35\right)\)
\(\left(2^{24}\right)^{83}\equiv1^{83}\equiv1\left(mod35\right)\)
\(\Rightarrow2^{1992}\cdot2^7\equiv1\cdot23\equiv23\left(mod35\right)\)
Vậy 21999 chia 35 dư 23