4 digits

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

If A , B , C , D A, B, C, D are four distinct digits, how many possible values does A B C D \overline{ABCD} have?

9 10 4 5 8 7 3 6

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1 solution

Michael Huang
Jul 30, 2017

A B C D D A B C C D A B + 0 C D A C C C C B \begin{array}{ c c c c c c c c c } & & A & B& C& D\\ & & D & A& B& C\\ & & C &D & A & B\\ + & & {\color{#D61F06}0} & C & D & A \\ \hline & C & C & C & C & B \end{array}

Observe the cryptarithm carefully that the first four terms in the sum list some of the possible arrangements of A , B , C A, B, C and D D , where the thousands digit of C D A \overline{CDA} is 'invisibly' B B . Noting that B = 0 B = 0 , we have A + C + D 0 m o d 10 A + C + D \equiv 0 \bmod 10 Since A + C + D 10 A + C + D \geq 10 , and A + C + D A + C + D exists in three remaining decimal places, there are 4 4 leftmost carry digits of the same value C C , which is nonzero. So either C = 1 C = 1 or C = 2 C = 2 .

  • If C = 2 C = 2 , then D + A 8 m o d 10 D + A \equiv 8 \bmod 10 . Since the carry digit is 2 2 , then D = A = 9 D = A = 9 , which is not possible.
  • If C = 1 C = 1 , then A + D 9 m o d 10 A + D \equiv 9 \bmod 10 , so A + D = 9 A + D = 9 . From here, we have 8 8 distinct integers from 2 2 to 9 9 . Since 1 1 and 0 0 are already used, this eliminates the candidates 8 8 and 9 9 , which then reduces to 6 6 integers from 2 2 to 7 7 . In this case, there are 7 2 + 1 = 6 7 - 2 + 1 = 6 possible ways to express two integers as the sum of 9 9 .

Thus, since the sum depends on the variables A A and B B , there are 1 1 6 = 6 1 \cdot 1 \cdot 6 = \boxed{6} possible choices to substitute A B C D \overline{ABCD} .


Note: I will check to see if there are other accurate/complete approaches for this problem. I believe my solution seems to be less rigorous because there can be other choices of B B that can also work.

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