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

(Fixed solution)
(→‎Solution: simplified - don't need a paragraph to reject A+B = D+10)
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==Solution==
 
==Solution==
−
Note from the addition of the last digits that <math>A+B=D\text{ or }D+10</math>. Assume the latter case is true; then we must have that <math>1+C+D=D\text{ or }10+D</math>, implying that <math>C=9</math>. In the addition of the third digits, we then have <math>1+B+A=D\text{ or }D+10</math>, a contradiction from our assumption that <math>A+B=D+10</math>. Thus <math>A+B=D</math>.
+
Note from the addition of the last digits that <math>A+B=D\text{ or }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.  
 +
Therefore <math>A+B=D</math>.  
  
−
This then implies that <math>C+D=D</math>, or <math>C=0</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>. Thus, if we have some <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>. Thus we have that <math>3\le D\le9</math>, so the number of possible values is <math>\boxed{\textbf{(C) }7}</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>.  
 +
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>.  
 +
Thus we have that <math>3\le D\le9</math>, so the number of possible values is <math>\boxed{\textbf{(C) }7}</math>
  
 
==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}}

Revision as of 00:55, 21 February 2014

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 $A+B=D\text{ or }D+10$. 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 $B+A=D$. Thus, if we have any $A,B,D$ such that $A+B=D$ then the addition will be satisfied. 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 $\boxed{\textbf{(C) }7}$

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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