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

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== Problem ==
 
== Problem ==
  
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== Solution ==
 
== Solution ==
  
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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\text{(B)}</math>.
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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>.
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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.
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-Edited slightly by RealityWrites
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2009|ab=A|num-b=8|num-a=10}}
 
{{AMC10 box|year=2009|ab=A|num-b=8|num-a=10}}
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{{MAA Notice}}

Revision as of 11:51, 8 September 2021

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*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

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