$A=1+3+3^2+3^3+...+3^{50}$
$=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^{48}+3^{49}+3^{50})$
$=(1+3+3^2)+3^3(1+3+3^2)+...+3^{48}(1+3+3^2)$
$=(1+3+3^2)(1+3^3+...+3^{48})$
$=13.(1+3^3+...+3^{48})$
Vì $13 \vdots 13$ nên $13.(1+3^3+...+3^{48}) \vdots 13$
Vậy $A \vdots 13$
$B=7+7^2+7^3+...+7^60$
$=(7+7^2+7^3)+...+(7^{58}+7^{59}+7^{60})$
$=7(1+7+7^2)+...+7^{58}(1+7+7^2)$
$=(7+...+7^{58})(8+49)$
$=57.(7+7^{58})$
Vì $57 \vdots 19$ nên $=57.(7+7^{58}) \vdots 19$
Vậy $B \vdots 19$
$C=1+3+3^2+3^3+...+3^{1991}$
$=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^{1989}+3^{1990}+3^{1991})$
$=(1+3+3^2)+3^3(1+3+3^2)+...+3^{1989}(1+3+3^2)$
$=(1+3+3^2)(1+3^3+...+3^{1989})$
$=13.(1+3^3+...+3^{1989})$
Vì $13 \vdots 13$ nên $13.(1+3^3+...+3^{1989}) \vdots 13$
Vậy $C \vdots 13$
Chúc em học tốt @@
