Difference between revisions of "2009 AMC 10A Problems/Problem 9"
Hashtagmath (talk | contribs) |
m (→Solution) |
||
Line 20: | Line 20: | ||
The prime factorization of <math>2009</math> is <math>2009 = 7\cdot 7\cdot 41</math>. As <math>a<b<2009</math>, the ratio must be positive and larger than <math>1</math>, hence there is only one possibility: the ratio must be <math>7</math>, and then <math>b=7\cdot 41</math>, and <math>a=41\Rightarrow\fbox{B}</math>. | The prime factorization of <math>2009</math> is <math>2009 = 7\cdot 7\cdot 41</math>. As <math>a<b<2009</math>, the ratio must be positive and larger than <math>1</math>, hence there is only one possibility: the ratio must be <math>7</math>, and then <math>b=7\cdot 41</math>, and <math>a=41\Rightarrow\fbox{B}</math>. | ||
+ | |||
+ | We know that this is important because the complete equation would be a*x^2=2009, and the only possible outcome for "x" is 7. | ||
+ | -Edited slightly by RealityWrites | ||
== See Also == | == See Also == |
Latest revision as of 11:51, 8 September 2021
Contents
Problem
Positive integers , , and , with , form a geometric sequence with an integer ratio. What is ?
Solution
The prime factorization of is . As , the ratio must be positive and larger than , hence there is only one possibility: the ratio must be , and then , and .
We know that this is important because the complete equation would be a*x^2=2009, and the only possible outcome for "x" is 7. -Edited slightly by RealityWrites
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.