Difference between revisions of "2009 AMC 10A Problems/Problem 9"

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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 <math>a\cdot x^2=2009,</math> and the only possible outcome for <math>x</math> is 7.
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We know that this is important because the complete equation would be <math>a\cdot x^2=2009,</math> and the only possible outcome for <math>x</math> is <math>7.</math>
 
-Edited slightly by RealityWrites - minor <math>\LaTeX</math> edits by BS2012
 
-Edited slightly by RealityWrites - minor <math>\LaTeX</math> edits by BS2012
  

Latest revision as of 11:20, 25 October 2024

Problem

Positive integers $a$, $b$, and $2009$, with $a<b<2009$, form a geometric sequence with an integer ratio. What is $a$?

$\mathrm{(A)}\ 7 \qquad \mathrm{(B)}\ 41 \qquad \mathrm{(C)}\ 49 \qquad \mathrm{(D)}\ 289 \qquad \mathrm{(E)}\ 2009$

Solution

The prime factorization of $2009$ is $2009 = 7\cdot 7\cdot 41$. As $a<b<2009$, the ratio must be positive and larger than $1$, hence there is only one possibility: the ratio must be $7$, and then $b=7\cdot 41$, and $a=41\Rightarrow\fbox{B}$.

We know that this is important because the complete equation would be $a\cdot x^2=2009,$ and the only possible outcome for $x$ is $7.$ -Edited slightly by RealityWrites - minor $\LaTeX$ edits by BS2012

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
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

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