Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
Line 17: Line 17:
  
 
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>. 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>.  
+
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>
−
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>
 
  
 
==Video Solution==
 
==Video Solution==

Revision as of 12:33, 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 $9$. 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_10&oldid=160093"