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2014 AMC 12B Problem 8

Problem 8 of 25EasierNumber Theory

In the addition shown below A,A, B,B, C,C, and DD are distinct digits. How many different values are possible for D?D? ABBCB+BCADADBDDD\begin{array}{cccccc} & A & B & B & C & B \\ + & B & C & A & D & A \\ \hline & D & B & D & D & D \end{array}

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Solution

The leftmost column shows A+B=DA+B = D with no carry out, so A+B9.A+B \le 9. Examining the tens and thousands columns (each of the form C+digit+carryC + \text{digit} + \text{carry} producing the same digit) forces C=0C = 0 and eliminates all carries. Every column then reduces to A+B=D,A+B = D, with A,B,C=0A, B, C=0 distinct. Since AA and BB are distinct positive digits, D=A+BD = A+B can be any value from 33 up to 9,9, giving 77 possibilities, for example (A,B,C,D)=(1,2,0,3),(A,B,C,D) = (1,2,0,3), (1,3,0,4),,(1,3,0,4), \ldots, (2,7,0,9).(2,7,0,9). Thus, the correct answer is C.

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Concepts: cryptarithm · place value · casework

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.