Question 995336
Let's go through all the choices


A) 0


If C = 0, then B+B ends with a zero and this is only possible if B = 0 or B = 5. 


if B = 0, then A,B,C all must be 0 but they are all different and nonzero


if B = 5, then it's not possible that B+B yields the same digit as B. So C = 0 is not possible. 


This rules out choice A.


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B) 6


If C = 6, then B has to be 3 or 8. 


If B = 3, then B+B=3+3 = 6 but that's not equal to B = 3. So B can't be 3.
If B = 8, then B+B=8+8 = 16. The units digit 6 is not equal to B = 8 (even with a carry). So B can't be 8.


This rules out choice B


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C) 8


If C = 8, then B has to be 4 or 9


If B = 4, then B+B doesn't have the units digit of B. So B = 4 is eliminated in this sub-case.
If B = 9, then B+B adds to 9+9 = 18 and we have a carry of 1 from the previous addition. So we really have 19. This makes B+B have a units digit of B. So it fits


So far, we see that it's possible for C = 8 and B = 9


That would make A = 1


Summary:
A = 1
B = 9
C = 8


Notice how BB+BB = ABC turns into 99+99 = 198 which fits the summary shown above. 


Final Answer: <font color="red">C) 8</font>


Note: we don't need to check D) 9 since we found the answer in choice C.