Question 995336: In the correct addition shown at the right, A,B, and C are different non-zero digits. What is the value of C?
BB
+BB
-------
ABC
A)0
B)6
C)8
D)9
Found 2 solutions by jim_thompson5910, MathLover1:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 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.
------------------------------------------------------------------
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
-------------------------------------------------------------------
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: C) 8
Note: we don't need to check D) 9 since we found the answer in choice C.
Answer by MathLover1(20850)  (Show Source):
|
|
|
