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

(New page: == Problem == Positive integers <math>a</math>, <math>b</math>, and <math>2009</math>, with <math>a<b<2009</math>, form a geometric sequence with an integer ratio. What is <math>a</math>?...)
 
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== Solution ==
 
== Solution ==
  
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=\boxed{41}</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\text{(B)}</math>.
  
 
== 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}}

Revision as of 20:23, 2 November 2012

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 $a=41\Rightarrow\text{(B)}$.

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AMC 10 Problems and Solutions
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