2014 AMC 10B Problems/Problem 10

Problem

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

$\begin{array}[t]{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}$

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

Solution

Note from the addition of the last digits that $A+B=D\text{ or }A+B=D+10$ . From the addition of the frontmost digits, $A+B$ cannot have a carry, since the answer is still a five-digit number. Also $A + B$ cant have a carry since then for the second column, $C + 1 + D$ cant equal $D$ . Therefore $A+B=D$ .

Using the second or fourth column, this then implies that $C=0$ , so that $B+C=B$ and $C+D=D$ . Note that all of the remaining equalities are now satisfied: $A+B=D, B+C=B,$ and $B+A=D$ . Since the digits must be distinct, the smallest possible value of $D$ is $1+2=3$ , and the largest possible value is $9$ . Thus we have that $3\le D\le9$ , so the number of possible values is $\boxed{\textbf{(C) }7}$

Video Solution (CREATIVE THINKING)

https://youtu.be/x5e1DJSXFss

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Video Solution

https://youtu.be/CCOjtLn2AKM

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2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 10 Problems and Solutions

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2014 AMC 10B Problems/Problem 10

Contents

Problem

Solution

Video Solution (CREATIVE THINKING)

Video Solution

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