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Difference between revisions of "2009 AMC 8 Problems/Problem 24"

(Created page with "==Problem== The letters <math>A</math>, <math>B</math>, <math>C</math> and <math>D</math> represent digits. If <math> \begin{tabular}{ccc}&A&B\\ +&C&A\\ \hline &D&A\end{tabular...")
 
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==Solution==
 
==Solution==
  
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Because <math> B+A=A </math>, <math>B</math> must be <math>0</math>
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Because <math> B+A=A </math>, <math>B</math> must be <math>0</math>.
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Next, because <math> B-A=A\implies0-A=A,</math> we get <math>A=5</math>
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Next, because <math> B-A=A\implies0-A=A,</math> we get <math>A=5</math> as the "0" mentioned above is actually 10 in this case.
  
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Now we can rewrite this as <math> \begin{tabular}{ccc}&5&0\\ +&C&5\\ \hline &D&5\end{tabular} </math>. Therefore, <math>D=5+C</math>
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Now we can rewrite <math> \begin{tabular}{ccc}&A&0\\ +&C&A\\ \hline &D&A\end{tabular} </math> as <math> \begin{tabular}{ccc}&5&0\\ +&C&5\\ \hline &D&5\end{tabular} </math>. Therefore, <math>D=5+C.</math>
  
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Finally, <math> A-1-C=0\implies{A=C+1}\implies{C=4} </math>, So we have <math> D=5+C\implies{D=5+4}=\boxed{9} </math>
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Finally, <math> A-1-C=0\implies{A=C+1}\implies{C=4} </math>, So we have <math> D=5+C\implies{D=5+4}=\boxed{\textbf{(E)}\ 9 } </math>.
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==See Also==
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{{AMC8 box|year=2009|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 12:07, 20 November 2024

Problem

The letters $A$, $B$, $C$ and $D$ represent digits. If $\begin{tabular}{ccc}&A&B\\ +&C&A\\ \hline &D&A\end{tabular}$and $\begin{tabular}{ccc}&A&B\\ -&C&A\\ \hline &&A\end{tabular}$,what digit does $D$ represent?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution

Because $B+A=A$, $B$ must be $0$. Next, because $B-A=A\implies0-A=A,$ we get $A=5$ as the "0" mentioned above is actually 10 in this case.

Now we can rewrite $\begin{tabular}{ccc}&A&0\\ +&C&A\\ \hline &D&A\end{tabular}$ as $\begin{tabular}{ccc}&5&0\\ +&C&5\\ \hline &D&5\end{tabular}$. Therefore, $D=5+C.$

Finally, $A-1-C=0\implies{A=C+1}\implies{C=4}$, So we have $D=5+C\implies{D=5+4}=\boxed{\textbf{(E)}\ 9 }$.

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 AJHSME/AMC 8 Problems and Solutions

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