2014 AMC 12B Problems/Problem 8

Problem

In the addition shown below $A$ , $B$ , $C$ , and $D$ are distinct digits. How many different values are possible for $D$ ?

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

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution

From the first column, we see $A+B < 10$ because it yields a single digit answer. From the fourth column, we see that $C+D$ equals $D$ and therefore $C = 0$ . We know that $A+B = D$ . Therefore, the number of values $D$ can take is equal to the number of possible sums less than $10$ that can be formed by adding two distinct natural numbers. Letting $A=1$ , and letting $B=2,3,4,5,6,7,8$ , we have $D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}$

Solution (Equation Algorithm)

It is intuitively obvious, even to the most casual observer that the problem statement can be rewritten as:

$10^4A + 10^4B + 10^3B + 10^3C + 10^2B + 10^2A + 10C + 10D + B + A = 10^4D + 10^3B + 10^2D + 10D + D$ . This equation can be simplified into:

$10^4A + 10^4B + 10^3C + 10^2B + 10^2A + 10C + B + A = 10^4D + 10^2D + D$ .

Now from here, it should hopefully make sense that $A + B = D$ by looking at the one's digit of both equations. Factoring out $A + B$ gives:

$10^4(A+B) + 10^3C + 10^2(A+B) + 10C + (B + A) = 10^4D + 10^2D + D$ .

Which equals: $10^4(D) + 10^3C + 10^2(D) + 10C + D = 10^4D + 10^2D + D$ .

This simplifies into: $10^3C + 10C = 0$ .

Therefore $C = 0$ .

This means that $A + B = D$ and $D < 10$ or else there would be parts carried over in the equation. The positive integers that satisfy this equation are a minimum $(2, 1)$ and a maximum of $(4, 5)$ . This means that $D = 3$ $, 4$ $, 5$ $, 6$ $, 7$ $, 8$ $, 9$ . Giving $\boxed{\textbf{(C)}\ 7}$

~PeterDoesPhysics

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2014 AMC 12B Problems/Problem 8

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Problem

Solution

Solution (Equation Algorithm)

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