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

Revision as of 11:14, 25 October 2024

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

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

yes

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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 == Solution ==
+yes
-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 <math>7.</math>
--Edited slightly by RealityWrites - minor <math>\LaTeX</math> edits by BS2012
 == See Also ==