\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{28}+3^{29}+3^{30}\right)\)

\(=\left(3+3^2+3^3\right)+3^3.\left(3+3^2+3^3\right)+...+3^{27}\left(3+3^2+3^3\right)\)

\(=39+3^3.39+...+3^{27}.39\)

Từ đây suy ra $A$ chia hết cho $13$ do $39$ chia hết cho $13$.