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

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If n and y are positive integers and 450y=n^3, which of the [#permalink]

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New post 27 Nov 2007, 11:30
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If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y / (3*4*5)
II. y / (9*2*5)
III. y / (3*2*25)
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New post 27 Nov 2007, 12:55
only I

450=3*5*3*2*5=2*3^2*5^2

y=k*2^2*3*5 where k - positive integer
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Re: PS - integer [#permalink]

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New post 27 Nov 2007, 12:56
tennisgurun wrote:
If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y / (3*4*5)
II. y / (9*2*5)
III. y / (3*2*25)



450Y = 5^2* 2*3^2 Y = N^3 , THEN Y MUST HAVE AT LEAST A FACTOR OF 5 AND 2^2 AND A 3

A is my answer
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Re: PS - integer [#permalink]

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New post 27 Nov 2007, 15:25
tennisgurun wrote:
If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y / (3*4*5)
II. y / (9*2*5)
III. y / (3*2*25)



A by testing the answers.

since n and y are + integers. n^3 is a perfect cube. So find the answer choice that when multiplied by 450 is a perfect cube.

Only I works: 3*4*5=60 --> 6*45=270 Then just add 2 zeros.

II and III do not work.
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New post 27 Nov 2007, 19:23
walker wrote:
only I

450=3*5*3*2*5=2*3^2*5^2

y=k*2^2*3*5 where k - positive integer


can someone explain the logic here ? i mean, i can break 450 down into primes, but how do we determine that 1 is the right answer ?
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New post 27 Nov 2007, 21:10
pmenon wrote:
can someone explain the logic here ? i mean, i can break 450 down into primes, but how do we determine that 1 is the right answer ?


for example II:

z= y / (9*2*5) = k*2^2*3*5 / (3*3*2*5)=k*2/3 ==> k=3: z-integer, but k=4: z-fraction. Therefore, II is insuff
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Re: PS - integer [#permalink]

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New post 27 Nov 2007, 22:15
tennisgurun wrote:
If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y / (3*4*5)
II. y / (9*2*5)
III. y / (3*2*25)


n^3 = 450 y
n^3 = 2 * 3^2 * 5^2 * y
Since n is an integer, the cube root of the right side of above equation should be an integer as well. The simplest way to make it an integer is if y = 2^2 * 3 * 5 = 60.

Now divide 60 by I II & III to see which work. Only I does so answer is A.
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Re: PS - integer [#permalink]

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New post 16 Dec 2007, 10:18
GK_Gmat wrote:
tennisgurun wrote:
If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y / (3*4*5)
II. y / (9*2*5)
III. y / (3*2*25)


n^3 = 450 y
n^3 = 2 * 3^2 * 5^2 * y
Since n is an integer, the cube root of the right side of above equation should be an integer as well. The simplest way to make it an integer is if y = 2^2 * 3 * 5 = 60.

Now divide 60 by I II & III to see which work. Only I does so answer is A.


i like this approach very much. thanks a lot for enlightening me :)
Re: PS - integer   [#permalink] 16 Dec 2007, 10:18
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If n and y are positive integers and 450y=n^3, which of the

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