Ta có :
\(5^{70}=\left(5^2\right)^{35}=25^{35}=\left(12.2+1\right)^{35}\equiv1\left(mod12\right)\)
\(7^{70}=\left(7^2\right)^{35}=49^{35}=\left(12.4+1\right)^{35}\equiv1\left(mod12\right)\)
\(\Rightarrow5^{70}+7^{50}\equiv2\left(mod12\right)\) hay \(5^{70}+7^{50}\) chia 12 dư 2