2014 AMC 10B Problem 10

Work column by column from the right, tracking carries.

Units: $B + A$ ends in $D$; call the carry $c_1$.

Tens: $C + D + c_1$ ends in $D$, so $C + c_1$ is $0$ or $10$. Either $C = 0$ and $c_1 = 0$, or $C = 9$ and $c_1 = 1$.

If $C = 9$, $c_1 = 1$: then $A + B = D + 10$ and the tens column produces a carry of $1$. The hundreds column reads $B + A + 1 = D + 11$, which ends in $D + 1$, not $D$. Impossible.

If $C = 0$, $c_1 = 0$: then $A + B = D$ with no carry, the tens column is $0 + D = D$ with no carry, the hundreds column is $B + A = D$ again with no carry, the thousands column is $B + 0 = B$, and the ten-thousands column is $A + B = D$. Everything is consistent.

So the only requirement is $D = A + B$ where $A$, $B$ are distinct digits, both nonzero (they differ from $C = 0$), and $D \le 9$. Since $A \ne B$, the sum $A + B$ automatically differs from both. The possible sums range from $1 + 2 = 3$ up to $9$, and every value $3, 4, \dots, 9$ is achievable (for example $1 + 2, 1 + 3, 1 + 4, 1 + 5, 1 + 6, 1 + 7, 1 + 8$).

That is $7$ values, so the answer is $\boxed{\textbf{(C)}\ 7}$.

In a cryptarithm, columns where the same letter appears in a summand and the sum (here $D$ in the tens place, $B$ in the thousands place) are the most informative: they force the other addend plus the carry to be $0$ or $10$. Resolving those columns first collapses the whole problem to one simple equation.

• Forgetting that D = A + B needs two distinct nonzero digits, so D = 1 and D = 2 are impossible; answering 8 or 9.
• Ignoring the $C = 9$ branch entirely (correct by luck) or, worse, accepting it and adding extra values of $D$.