\(2^{2011}:31\) =x dư ?
như vầy ai chơi?
Ta có:
\(2^5\equiv1\left(mod31\right)\)
\(\Rightarrow2^{2010}\equiv1\left(mod31\right)\)
\(\Rightarrow2^{2011}\equiv2\left(mod31\right)\)
\(\Rightarrow2^{2011}-2\equiv0\left(mod31\right)\)
\(\Rightarrow2^{2011}-2⋮31\)
\(\Rightarrow2^{2011}\) chia 31 dư 2.
Chúc bạn học tốt!