Bài giải :

$A=1+3+{{3}^{2}}+...+{{3}^{11}}$

$A=\left( 1+3+{{3}^{2}}+{{3}^{3}} \right)+\left( {{3}^{4}}+...+{{3}^{7}} \right)+\left( {{3}^{8}}+...+{{3}^{11}} \right)$

$A=\left( 1+3+{{3}^{2}}+{{3}^{3}} \right)+{{3}^{4}}\left( 1+3+{{3}^{2}}+{{3}^{3}} \right)+{{3}^{8}}\left( 1+3+{{3}^{2}}+{{3}^{3}} \right)$

$A=40+{{3}^{4}}.40+{{3}^{8}}.40$

$A=40\left( 1+{{3}^{4}}+{{3}^{8}} \right)$

Vậy A có chia hết cho 40.