A number divisible by both $$6$$ and $$20$$ must also be divisible by. 
Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Divisibility by $$6$$ \& $$20$$ does not promise divisibility by $$7$$, $$13$$, or $$8$$.  $$\rm A$$. $$12 = 4\times3$$; $$\rm B$$. $$14 = 2\times7$$; $$\rm C$$. $$26 = 2\times13$$; $$\rm D$$. $$120 = 8\times15$$. 