Difference between revisions of "2009 AMC 8 Problems/Problem 24"
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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 | + | 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 , , and represent digits. If and ,what digit does represent?
Solution
Because , must be . Next, because we get as the "0" mentioned above is actually 10 in this case.
Now we can rewrite as . Therefore,
Finally, , So we have .
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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