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

m (Solution)
m (Solution)
Line 13: Line 13:
 
==Solution==
 
==Solution==
 
Note from the addition of the last digits that <math>A+B=D\text{ or }A+B=D+10</math>.  
 
Note from the addition of the last digits that <math>A+B=D\text{ or }A+B=D+10</math>.  
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 A + B cant have a carry since then for the second column, C + 1 + D cant equal D.  
+
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 D.  
 
Therefore <math>A+B=D</math>.  
 
Therefore <math>A+B=D</math>.  
  

Revision as of 16:36, 20 October 2022

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

https://youtu.be/CCOjtLn2AKM

~savannahsolver

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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