2014 AMC 10B Problem 10 / AMC 12B Problem 8

2014 AMC 10B 10 / AMC 12B 8. In the addition below ABC, 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. A + B < 10.

A + B 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 C + D = D, 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 A + B < 10, can we find any other constraints on the values of AB, and (most importantly D)? As long as A + B < 10, 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 “ABC, and D are distinct digits”. D can be 2 only if A = B = 1, 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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One Response to 2014 AMC 10B Problem 10 / AMC 12B Problem 8

  1. Pingback: 2014 Singapore MO (Junior Rd 1) Problem 7 | Beyond Solutions

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