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If x and y are positive integers 4500x = y3, what is the

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If x and y are positive integers 4500x = y3, what is the [#permalink] New post 26 Jun 2007, 21:22
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If x and y are positive integers 4500x = y3, what is the minimum possible value of x?

(A) 5
(B) 6
(C) 12
(D) 30
(E) 90

No solution is available. I reasoned this out and think I got the correct answer. Please explain your solution.
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Re: PS Question I found somewhere [#permalink] New post 26 Jun 2007, 21:52
ggarr wrote:
If x and y are positive integers 4500x = y3, what is the minimum possible value of x?

(A) 5
(B) 6
(C) 12
(D) 30
(E) 90

No solution is available. I reasoned this out and think I got the correct answer. Please explain your solution.


should be B.

y^3 = 4500x
y^3 = (30) (30) (5x)

so x should be 6 to make 5x = 30.
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 [#permalink] New post 26 Jun 2007, 22:30
prime factorization of 4500 = 5x5x5x3x3x2x2 - so it needs 2x3 to make it a perfect cube

2x3 =6
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 [#permalink] New post 27 Jun 2007, 09:12
4500 = 2^2 * 3^2 * 5*3

we need 2*3 to make it (2*3*5)^3

x = 6.
B.
  [#permalink] 27 Jun 2007, 09:12
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