2009 AMC 8 Problems/Problem 24

Revision as of 03:30, 31 October 2020 by Penguin spellcaster (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS