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Difference between revisions of "2014 AMC 10B Problems/Problem 10"

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<math> \textbf {(A) } 2 \qquad \textbf {(B) } 4 \qquad \textbf {(C) } 7 \qquad \textbf {(D) } 8 \qquad \textbf {(E) } 9 </math>
  
 
==Solution==
 
==Solution==
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Note from the addition of the last digits that <math>A+B=D\text{ or }A+B=D+10</math>.
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From the addition of the frontmost digits, <math>A+B</math> cannot have a carry, since the answer is still a five-digit number. Also <math>A + B</math> cant have a carry since then for the second column, <math>C + 1 + D</math> cant equal <math>D</math>.
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Therefore <math>A+B=D</math>.
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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>.
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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 we have that <math>3\le D\le9</math>, so the number of possible values is <math>\boxed{\textbf{(C) }7}</math>
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/x5e1DJSXFss
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/CCOjtLn2AKM
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~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2014|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:01, 2 July 2023

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

~Education, the Study of Everything


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

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