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If the four-digit number $$28X8$$ is divisible by $$3$$, how many possible values are there for $$X$$? 
Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
$$2+8+8=18$$, $$18$$ is multiple of $$3$$  Thus, $$X$$ itself must be a multiple of $$3$$, or $$0$$.  We have $$3, 6, 9$$ and $$0$$. Four possible value. 