Skip to main content

2014 AMC 10B Problem 10

Problem 10 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}{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}

Answer choices

Show solution

Solution

From the leftmost column, there is no carry into a sixth digit, so A+B=D9A+B=D\le 9. The units column is B+A=DB+A=D, so it also has no carry. The tens column then gives C+D=DC+D=D, hence C=0C=0. Since AA and BB are distinct nonzero digits, D=A+BD=A+B can be any digit from 33 through 99. For example, (A,B)=(1,2),(A,B)=(1,2), (1,3),(1,3), (2,3),(2,3), (2,4),(2,4), (2,5),(2,5), (2,6),(2,6), (2,7)(2,7) give D=3,4,5,6,7,8,9D=3,4,5,6,7,8,9. Thus there are 77 possible values of DD, and the correct answer is C .

More practice

Concepts: cryptarithm · digits · casework

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