2014 AMC 12B Problem 8

Question

In the addition shown below

A,

B,

C,

and

D

are distinct digits. How many different values are possible for

D?

\begin{array}{cccccc} & A & B & B & C & B \\ + & B & C & A & D & A \\ \hline & D & B & D & D & D \end{array}

2

4

7

8

9

The leftmost column gives

A+B=D

with no carry, so

A+B \le 9

The second and fourth columns force

C=0

and no carries anywhere

The leftmost column shows

A+B = D

with no carry out, so

A+B \le 9.

Examining the tens and thousands columns (each of the form

C + \text{digit} + \text{carry}

producing the same digit) forces

C = 0

and eliminates all carries.

Every column then reduces to

A+B = D,

with

A, B, C=0

distinct. Since

A

and

B

are distinct positive digits,

D = A+B

can be any value from

3

up to

9,

giving

7

possibilities, for example

(A,B,C,D) = (1,2,0,3),

(1,3,0,4), \ldots,

(2,7,0,9).

Thus, the correct answer is C.

Problem 8 in other years