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2009 AMC 10A Problems/Problem 9

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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 $a=\boxed{41}$ .

See Also

2009 AMC 10A (Problems • 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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2009_AMC_10A_Problems/Problem_9"

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