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

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m (Solution)
 
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Because <math> B+A=A </math>, <math>B</math> must be <math>0</math>.
 
Because <math> B+A=A </math>, <math>B</math> must be <math>0</math>.
Next, because <math> B-A=A\implies0-A=A,</math> we get <math>A=5</math>
+
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.
  
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>
+
Now we can rewrite <math> \begin{tabular}{ccc}&A&B\\ +&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>
  
 
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>.
 
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>.

Latest revision as of 03:30, 31 October 2020

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 $\begin{tabular}{ccc}&A&B\\ +&C&A\\ \hline &D&A\end{tabular}$ 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 (ProblemsAnswer KeyResources)
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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