Ta có $B={{3}^{2022}}+{{2}^{2022}}+{{3}^{2020}}+{{2}^{2020}}$
$=\left( {{3}^{2022}}+{{3}^{2020}} \right)+\left( {{2}^{2022}}+{{2}^{2020}} \right)$
$={{3}^{2020}}\left( {{3}^{2}}+1 \right)+{{2}^{2019}}\left( {{2}^{3}}+2 \right)$
$={{3}^{2020}}.10+{{2}^{2019}}.10$
$=10.\left( {{3}^{2020}}+{{2}^{2019}} \right)$
Vì $10.\left( {{3}^{2020}}+{{2}^{2019}} \right)\vdots 10$ nên $B\vdots 10$