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Difference between revisions of "2014 AMC 12B Problems/Problem 8"
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From the first column, we see <math>A+B < 10</math> because it yields a single digit answer.  From the fourth column, we see that <math>C+D</math> equals <math>D</math> and therefore <math>C = 0</math>.  We know that <math>A+B = D</math>.  Therefore, the number of values <math>D</math> can take is equal to the number of possible sums less than <math>10</math> that can be formed by adding two distinct natural numbers.  Letting <math>A=1</math>, and letting <math>B=2,3,4,5,6,7,8</math>, we have  
 
From the first column, we see <math>A+B < 10</math> because it yields a single digit answer.  From the fourth column, we see that <math>C+D</math> equals <math>D</math> and therefore <math>C = 0</math>.  We know that <math>A+B = D</math>.  Therefore, the number of values <math>D</math> can take is equal to the number of possible sums less than <math>10</math> that can be formed by adding two distinct natural numbers.  Letting <math>A=1</math>, and letting <math>B=2,3,4,5,6,7,8</math>, we have  
 
<cmath>D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}</cmath>
 
<cmath>D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}</cmath>
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{{AMC12 box|year=2014|ab=B|num-b=7|num-a=9}}
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{{MAA Notice}}

Revision as of 12:09, 21 February 2014

Problem

In the addition shown below $A$, $B$, $C$, and $D$ are distinct digits. How many different values are possible for $D$?

\[\begin{tabular}{cccccc}&A&B&B&C&B\\ +&B&C&A&D&A\\ \hline &D&B&D&D&D\end{tabular}\]

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution

From the first column, we see $A+B < 10$ because it yields a single digit answer. From the fourth column, we see that $C+D$ equals $D$ and therefore $C = 0$. We know that $A+B = D$. Therefore, the number of values $D$ can take is equal to the number of possible sums less than $10$ that can be formed by adding two distinct natural numbers. Letting $A=1$, and letting $B=2,3,4,5,6,7,8$, we have \[D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}\]

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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All AMC 12 Problems and Solutions

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