**2014 AMC 10B 10 / AMC 12B 8.** In the addition below *A*, *B*, *C*, and *D* are distinct digits. How many different values are possible for *D*?

* A B B C B
+ B C A D A
—————-
*

*D B D D D*

(A) 2 (B) 4 (C) 7 (D) 8 (E) 9

For these sort of problems, the key is usually to * figure out when and where digits are “carried over”*. A good place to start is by

*.*

**looking at the leftmost digits**In this problem, the number of digits of the sum (*DBDDD*) is the same as the number of digits each operand has (*ABBCB *and *BCADA)*. This means that there is no “carry over” in the leftmost column, i.e.

occurs 2 more times in the problem, in the middle and rightmost column. Since the rightmost column does not have any carry over, looking at the 2nd rightmost column we must have , which implies that *C* must be 0. With this information, we can substitute 0 for *C *to get this:

* A B B 0 B
+ B 0 A D A
—————-
*

*D B D D D*

Now, other than , can we find any other constraints on the values of *A*, *B*, and (most importantly *D*)? As long as , columns 1, 3 and 5 will take care of themselves. Columns 2 and 4 evidently will always be correct since there is no “carry over”.

Does that mean that *D* can take on any value from 0 to 9 (inclusive)? Well, no. *D* can’t be zero since it is the leading digit of the sum. For the same reason *A* and *B* can’t be zero, which means that *D* can’t be 1 either.

Can *D* be equal to 2? This is where a * close reading of the question is important*. The question states that “

*A*,

*B*,

*C*, and

*D*are

**distinct**digits”.

*D*can be 2 only if , which contradicts the question’s condition. Hence,

*D*cannot be 2.

It is quite easy to check that *D* can take on the rest of the values (3 to 9 inclusive). Hence, the answer is 7, or **(C)**.

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