a, Đặt : A \(=2^{9^{1945}}\)
Ta có :
\(2^3\equiv1\left(mod7\right)\); \(9\equiv0\left(mod3\right)\Rightarrow9^{1945}\equiv0\left(mod3\right)\)
Đặt : \(9^{1945}\)=3k ( k \(\in N\)
\(\Rightarrow A=2^{3k}=\left(2^3\right)^k=8^k\equiv1\left(mod7\right)\)
Vậy : A chia 7 dư 1
b, Đặt \(B=3^{2^{1930}}\)
Ta có : \(3^3\equiv-1\left(mod7\right);8\equiv-1\left(mod3\right)\)
\(B=\left(2^3\right)^{623}.2=2^{1930}\equiv-1.2\equiv-2\left(mod3\right)\equiv1\left(mod3\right)\)
=> \(2^{1930}-1=3k\left(k=2k+1\right)\Rightarrow3^{2^{1930}-1}=3^{3k}=27^k\equiv-1\left(mod7\right)\)
B=\(3.3^{2^{1930}-1}\equiv-1.3\left(mod7\right)\equiv4\left(mod7\right)\)
Vậy : B chia 7 dư 4
