Giải:
Ta có: A= 1+5+52+...+52018
\(\Leftrightarrow A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5+5^6\right)+\left(5^7+5^8+5^9+5^{10}\right)+...+\left(5^{2015}+5^{2016}+5^{2017}+5^{2018}\right)\)
\(\Leftrightarrow A=31+5^3\left(1+5+5^2+5^3\right)+...+5^{2015}\left(1+5+5^2+5^3\right)\)
\(\Leftrightarrow A=31+5^3.156+...+5^{2015}.156\)
\(\Leftrightarrow A=31+156.\left(5^3+...+5^{2015}\right)\)= 31+ 13.12.(53+...+52015)
Vì: \(\left\{{}\begin{matrix}13.12.\left(5^3+...+5^{2015}\right)⋮13\\31chia13du5\end{matrix}\right.\)
\(\Rightarrow156.\left(5^3+...+5^{2015}\right)+31\) chia 13 dư 5
\(\Leftrightarrow A\) chia 13 dư 5.
Vậy: A chia 13 dư 5
