Difference between revisions of "2009 AMC 8 Problems/Problem 24"
Sounderstid (talk | contribs) m |
|||
| (4 intermediate revisions by 4 users not shown) | |||
| Line 7: | Line 7: | ||
==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{\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>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=23|num-a=25}} | {{AMC8 box|year=2009|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 12:07, 20 November 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
