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2014 AMC 10B Problems/Problem 10: Difference between revisions

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Using the second or fourth column, this then implies that <math>C=0</math>, so that <math>B+C=B</math> and <math>C+D=D</math>.  
Using the second or fourth column, this then implies that <math>C=0</math>, so that <math>B+C=B</math> and <math>C+D=D</math>.  
Note that all of the remaining equalities are now satisfied: <math>A+B=D, B+C=B,</math> and <math>B+A=D</math>.  
Note that all of the remaining equalities are now satisfied: <math>A+B=D, B+C=B,</math> and <math>B+A=D</math>. Since the digits must be distinct, the smallest possible value of <math>D</math> is <math>1+2=3</math>, and the largest possible value is <math>9</math>.  
Thus, if we have any <math>A,B,D</math> such that <math>A+B=D</math> then the addition will be satisfied.
Since the digits must be distinct, the smallest possible value of <math>D</math> is <math>1+2=3</math>, and the largest possible value is <math>9</math>.  
Any of these values can be obtained by taking <math>A=1,B=D-1</math>.  
Any of these values can be obtained by taking <math>A=1,B=D-1</math>.  
Thus we have that <math>3\le D\le9</math>, so the number of possible values is <math>\boxed{\textbf{(C) }7}</math>
Thus we have that <math>3\le D\le9</math>, so the number of possible values is <math>\boxed{\textbf{(C) }7}</math>

Revision as of 12:32, 12 August 2021

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. 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 $9$. Any of these values can be obtained by taking $A=1,B=D-1$. Thus we have that $3\le D\le9$, so the number of possible values is $\boxed{\textbf{(C) }7}$

Video Solution

https://youtu.be/CCOjtLn2AKM

~savannahsolver

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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