Đặt \(S=1+5+5^2+5^3+5^4+...+5^9\)
\(\Leftrightarrow S=1+\left(5+5^2+5^3+5^4+...+5^9\right)\)
Đặt \(A=5+5^2+5^3+....+5^9\)
\(\Leftrightarrow A=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+\left(5^7+5^8+5^9\right)\)
\(\Leftrightarrow A=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+5^7\left(1+5+5^2\right)\)
\(\Leftrightarrow A=5\left(1+5+25\right)+5^4\left(1+5+25\right)+5^7\left(1+5+25\right)\)
\(\Leftrightarrow A=5\cdot31+5^4\cdot31+5^7\cdot31\)
\(\Leftrightarrow A=31\left(5+5^4+5^7\right)\)
=> A chia hết cho 31
Thay A=\(31\left(5+5^4+5^7\right)\)thay vào S ta có:
\(S=1+31\left(5+5^4+5^7\right)\)
=> S chia 31 dư 1
