$$2+8+8=18$$, $$18$$ 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. 
If the four-digit number $$28X8$$ is divisible by $$3$$, how many possible values are there for $$X$$? 
1: $$1$$ 
2: $$2$$ 
3: $$3$$ 
4: $$4$$ 
5: $$5$$ 