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# If y and n are positive integers and 450y = N^3 , which of

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Senior Manager
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If y and n are positive integers and 450y = N^3 , which of [#permalink]  15 Sep 2006, 16:02
If y and n are positive integers and 450y = N^3 , which of the following MUST be an integer:

I. y/(3 * 2^2 * 5)

II y/(3^2 * 2 * 5)

III y/(3 * 2 * 5^2)
Manager
Joined: 01 Jun 2006
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I found I the only answer
Because 450=5^2*3^2*2
So N must be divisable by 3*5*2 => N^3 divisable by 450*5*3*2^2
=> y divisable by 2^2*3*5
Senior Manager
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but they are all contain the common factors of 3*5*2 so why aren't the others valid?
Senior Manager
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I think I have it. y and N are both integers

450y = N^3/450

450 = 3^2 * 5^2 * 2

Substituting N^3/450 into each of the possibilities gives:

I N^3/ (3^2 * 5^2 * 2)(3 * 2^2 * 5) or N^3/3^3 * 5^3 * 2^3

This is the only answer choice that you can take the cube root of leaving an integer.
Senior Manager
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Given 450y= N^3
i.e 2 x3^2 x 5^2 x y is a perfect cube.
This will happen provided the minimum value of y is 2^2 x 3 x 5 i.e 60.
So the general value of Y is 60 x k^3
From the given options for any value of k option 1 is true.
So option 1 must be true.

Regards,
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Last edited by cicerone on 25 Sep 2008, 00:09, edited 1 time in total.
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If y and n are positive integers and 450y = N^3 , which of the following MUST be an integer:

I. y/(3 * 2^2 * 5)

II y/(3^2 * 2 * 5)

III y/(3 * 2 * 5^2)

THE PART IN READ MEANS THE LEAST POSSIBLE VALUE OF N^3 IT BECAME A FAMOUS GMAT TRICK THESE DAYS )

450 = 5^2*3^2*2

THUS Y MUST BE AT LEAST (5*3*2^2) THUS MY ANSWER IS 1
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Re: GMAT Prep Q11 [#permalink]  17 Sep 2006, 02:48
I agree with I, y must contain 60=2^2*5*3 for 450y to be a cube. the only option that divides 2^2 is I

If y and n are positive integers and 450y = N^3 , which of the following MUST be an integer:

I. y/(3 * 2^2 * 5)

II y/(3^2 * 2 * 5)

III y/(3 * 2 * 5^2)

_________________

________________________________
"Amicus Plato, sed magic amica veritas"

Re: GMAT Prep Q11   [#permalink] 17 Sep 2006, 02:48
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