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If x and y are positive integers and 180x=y^3...

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If x and y are positive integers and 180x=y^3... [#permalink] New post 04 Sep 2010, 10:34
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Question Stats:

69% (01:54) correct 31% (01:35) wrong based on 159 sessions
If x and y are positive integers and 180x=y^3, which of the following must be an integer?

I. x/(2^2*3*5)
II. x/(2*3^2*5)
III. x/(2*3*5^2)

(A) None
(B) I only
(C) II only
(D) III only
(E) I and II
[Reveal] Spoiler: OA
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Re: If x and y are positive integers and 180x=y^3... [#permalink] New post 04 Sep 2010, 11:01
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rraggio wrote:
If x and y are positive integers and 180x=y^3, which of the following must be an integer?

I. x/(2^2*3*5)
II. x/(2*3^2*5)
III. x/(2*3*5^2)

(A) None
(B) I only
(C) II only
(D) III only
(E) I and II


x and y are integers and 180x=y^3 --> 180x equals to cube of an integer (y^3). 180x=2^2*3^2*5*x=y^3. The smallest value of x for which 2^2*3^2*5*x is a cube of an integer is when x=2*3*5^2 (to complete powers of 2, 3 and 5 to cubes). In this case 180x=(2^2*3^2*5)*(2*3*5^2)=(2*3*5)^3=cube \ of \ an \ integer. So x is definitely divisible by 2*3*5^2.

Answer: D.

Similar problems:
number-properties-92562.html?hilit=following%20must%20none#p712584
og-quantitative-91750.html#p704028
division-factor-88388.html#p666722

Hope it helps.
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Re: If x and y are positive integers and 180x=y^3... [#permalink] New post 11 Jan 2014, 22:06
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Re: If x and y are positive integers and 180x=y^3...   [#permalink] 11 Jan 2014, 22:06
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