Chứng minh rằng \(A = 1 + 5 + {5^2} +  \ldots  + {5^{402}} + {5^{403}} + {5^{404}}\) chia hết cho \(31\).

a) Chứng minh rằng \(A = 1 + 5 + {5^2} +  \ldots  + {5^{402}} + {5^{403}} + {5^{404}}\) chia hết cho \(31\).

\(\begin{array}{l}A = 1 + 5 + {5^2} +  \ldots  + {5^{402}} + {5^{403}} + {5^{404}}\\A = \left( {1 + 5 + {5^2}} \right) + \left( {{5^3} + {5^4} + {5^5}} \right) +  \ldots  + \left( {{5^{402}} + {5^{403}} + {5^{404}}} \right)\\A = \left( {1 + 5 + {5^2}} \right) + {5^3}.\left( {1 + 5 + {5^2}} \right) +  \ldots  + {5^{402}}.\left( {1 + 5 + {5^2}} \right)\\A = \left( {1 + 5 + {5^2}} \right).\left( {1 + {5^3} +  \ldots  + {5^{402}}} \right)\\A = 31.\left( {1 + {5^3} +  \ldots  + {5^{402}}} \right)\,\, \vdots \,\,31\\ \Rightarrow A\,\, \vdots \,\,31\end{array}\)