Bài giải :

$C=1+3+{{3}^{2}}+{{3}^{3}}+...+{{3}^{119}}$

$C=\left( 1+3+{{3}^{2}} \right)+\left( {{3}^{3}}+{{3}^{4}}+{{3}^{5}} \right)+...+\left( {{3}^{117}}+{{3}^{118}}+{{3}^{119}} \right)$

$C=\left( 1+3+{{3}^{2}} \right)+{{3}^{3}}\left( 1+3+{{3}^{2}} \right)+...+{{3}^{117}}\left( 1+3+{{3}^{2}} \right)$

$C=\left( 1+{{3}^{3}}+...+{{3}^{117}} \right)\left( 1+3+{{3}^{2}} \right)$

$C=\left( 1+{{3}^{3}}+...+{{3}^{117}} \right).13$

Vậy C chia hết cho 13.