Difference between revisions of "2005 AMC 12A Problems/Problem 8"
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Let <math>A,M</math>, and <math>C</math> be digits with | Let <math>A,M</math>, and <math>C</math> be digits with | ||
| − | <cmath>(100A+10M+C)(A+M+C) = 2005</cmath> | + | <cmath>(100A+10M+C)(A+M+C) = 2005.</cmath> |
What is <math>A</math>? | What is <math>A</math>? | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:49, 1 July 2025
Problem
Let 
, and 
be digits with
![\[(100A+10M+C)(A+M+C) = 2005.\]](http://latex.artofproblemsolving.com/2/4/5/245fb2387e465e04164cd72cf11fe858db1a90b3.png)
What is 
?
Solution
Clearly the two quantities are both integers, so we check the prime factorization of 
. It is easy to see now that 
works, so the answer is 
.
See also
| 2005 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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 AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
