+ Chứng minh chia hết cho 10
$\begin{align}
& A=\left( 3+{{3}^{3}} \right)+\left( {{3}^{2}}+{{3}^{4}} \right)+\left( {{3}^{5}}+{{3}^{7}} \right)...+\left( {{3}^{2002}}+{{3}^{2004}} \right) \\
& \,\,\,\,\,=3\left( 1+{{3}^{2}} \right)+{{3}^{2}}\left( 1+{{3}^{2}} \right)+{{3}^{5}}\left( 1+{{3}^{2}} \right)...+{{3}^{2002}}\left( 1+{{3}^{2}} \right) \\
& \,\,\,\,\,=\left( 1+{{3}^{2}} \right)\left( 3+{{3}^{2}}+{{3}^{5}}+...+{{3}^{2002}} \right) \\
& \,\,\,\,\,=10\left( 3+{{3}^{2}}+{{3}^{5}}+...+{{3}^{2001}} \right) \\
\end{align}$
$\Rightarrow A\,\,\vdots \,\,10\,\,\,\,\,\left( 1 \right)$
+ Chứng minh chia hết cho 13
$\begin{align}
& A=\left( 3+{{3}^{2}}+{{3}^{3}} \right)+\left( {{3}^{4}}+{{3}^{5}}+{{3}^{6}} \right)+...+\left( {{3}^{2002}}+{{3}^{2003}}+{{3}^{2004}} \right) \\
& \,\,\,\,\,=3\left( 1+3+{{3}^{2}} \right)+{{3}^{4}}\left( 1+3+{{3}^{2}} \right)+...+{{3}^{2002}}\left( 1+3+{{3}^{2}} \right) \\
& \,\,\,\,\,=\left( 1+3+{{3}^{2}} \right)\left( 3+{{3}^{4}}+...+{{3}^{2002}} \right) \\
& \,\,\,\,\,=13\left( 3+{{3}^{4}}+...+{{3}^{2002}} \right) \\
\end{align}$
$\Rightarrow A\,\,\vdots \,\,13\,\,\,\,\,\left( 2 \right)$
Từ (1) và (2)
$\Rightarrow A\,\,$là bội chung của 13 và 10
$\Rightarrow A\,\,\vdots \,\,130$

thanks nha