Lời giải:
Ta có:
\(A=3+3^2+3^3+3^4+...+3^{25}\)
\(=3+(3^2+3^3+3^4+3^5)+(3^6+3^7+3^8+3^9)+...+(3^{22}+3^{23}+3^{24}+3^{25})\)
\(=3+3^2(1+3+3^2+3^3)+3^6(1+3+3^2+3^3)+....+3^{22}(1+3+3^2+3^3)\)
\(=3+(1+3+3^2+3^3)(3^2+3^6+...+3^{22})\)
\(=3+40(3^2+3^6+...+3^{22})\)
Từ đây suy ra $A$ chia $40$ dư $3$