Bài giải :

$A=3+{{3}^{2}}+{{3}^{3}}+...+{{3}^{60}}$

$A=\left( 3+{{3}^{2}}+{{3}^{3}} \right)+\left( {{3}^{4}}+{{3}^{5}}+{{3}^{6}} \right)+...+\left( {{3}^{58}}+{{3}^{59}}+{{3}^{60}} \right)$

$A=3\left( 1+3+{{3}^{2}} \right)+{{3}^{4}}\left( 1+3+{{3}^{2}} \right)+...+{{3}^{58}}\left( 1+3+{{3}^{2}} \right)$

$A=\left( 1+3+{{3}^{2}} \right)\left( 3+{{3}^{4}}+...+{{3}^{58}} \right)$

$A=13\left( 3+{{3}^{4}}+...+{{3}^{58}} \right)$

Vậy A có chia hết cho 13.