`B=3+3^2+3^3+...+3^{2009}+3^{2010}`
`B=3.(1+3+3^2)+...+3^{2008}(1+3+3^2)`
`B=(1+3+3^2).(3+3^4+...+3^{2008})`
`B=13.(3+3^4+...+3^{2008})`
Vì `13\vdots13` nên: `13.(3+3^4+...+3^{2008})\vdots13`
`=>B\vdots13`
Vậy `B=3+3^2+3^3+...+3^{2009}+3^{2010}\vdots13`