Difference between revisions of "2009 AMC 8 Problems/Problem 24"
Mrdavid445 (talk | contribs) (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== | ||
− | 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&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> |
− | 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> | + | 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>. |
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2009|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} |
Latest revision as of 23:39, 1 January 2024
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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