Ta có: $A=3+{{3}^{2}}+{{3}^{3}}+{{3}^{4}}+...+{{3}^{25}}$
$A=3+\left( {{3}^{2}}+{{3}^{3}}+{{3}^{4}}+{{3}^{5}} \right)+...+\left( {{3}^{22}}+{{3}^{23}}+{{3}^{24}}+{{3}^{25}} \right)$
$A=3+{{3}^{2}}\left( 1+3+{{3}^{2}}+{{3}^{3}} \right)+...+{{3}^{22}}\left( 1+3+{{3}^{2}}+{{3}^{3}} \right)$
$A=3+{{3}^{2}}.40+...+{{3}^{22}}.40$
$A=3+40.\left( {{3}^{2}}+{{3}^{6}}+...+{{3}^{22}} \right)$
Vậy A chia cho 40 có số dư là 3