Ta có: \(2730\equiv0\left(mod7\right)\Rightarrow1730^{10}\equiv0\left(mod7\right)\left(1\right)\)
\(927309\equiv5\left(mod7\right)\)
\(\Rightarrow927309^{10^2}\equiv5^{10^2}\left(mod7\right)\)
Mà \(5^6\equiv1\left(mod7\right)\)
\(\Rightarrow5^{100}=5^{96}.5^4\equiv5^4\equiv2\left(mod7\right)\)
\(\Rightarrow927309^{10^2}\equiv2\left(mod7\right)\left(2\right)\)
Ta lại có: \(27309\equiv2\left(mod7\right)\)
\(\Rightarrow27309^{10^n}\equiv2^{10^n}\left(mod7\right)\)
Mà \(2^{10^n}=2.2^{10^n-1}\equiv2\left(mod7\right)\left(3\right)\)
Từ (1), (2), (3) ta có
\(A=\left(2730^{10}+927309^{10^2}+27309^{10^3}+...+27309^{10^{10}}\right)\equiv\left(0+2+2+...+2\right)\equiv18\equiv4\left(mod7\right)\)
Vậy số dư của A cho 7 là 4
