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

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

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.

@@ Line 13: / Line 13: @@
 ==Solution==
 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.
+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>.
 Therefore <math>A+B=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 (CREATIVE THINKING)==
+https://youtu.be/x5e1DJSXFss
+~Education, the Study of Everything
 ==Video Solution==

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