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

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

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New post 09 Mar 2010, 01:50
2
00:00
A
B
C
D
E

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Question Stats:

86% (01:29) correct 14% (03:32) wrong based on 112 sessions

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If n and y are positive integers and 450y = n^3, which of the following must be an integer?

1 - y/(3*2^2*5)
2 - y/(3^2*2*5)
3 - y/(3*2*5^2)
4 - y/(5^3*2^2)
5 - y/(5*3*2)

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-n-and-y-are-positive-integers-and-450y-n-92562.html
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Re: 450y=n^3, finding integer  [#permalink]

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New post Updated on: 09 Mar 2010, 02:59
mustdoit wrote:
If n and y are positive integers and 450y = n^3, which of the following must be an integer?

1 - y/(3*2^2*5)
2 - y/(3^2*2*5)
3 - y/(3*2*5^2)
4 - y/(5^3*2^2)
5 - y/(5*3*2)


N is a perfect cube so it is factored to powers of three:

for Y to complement the prime factors of 450 , Y is divisable by 3 * 2^2 * 5. which is choice A.
However , by the same reasoning Y is also divisable by 5*3*2.

So both choices A and E work here unless i am missing something..

Originally posted by Pedros on 09 Mar 2010, 02:52.
Last edited by Pedros on 09 Mar 2010, 02:59, edited 1 time in total.
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Re: 450y=n^3, finding integer  [#permalink]

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New post 09 Mar 2010, 02:54
I would like to see the official explaination if you have it.
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Re: 450y=n^3, finding integer  [#permalink]

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New post 09 Mar 2010, 04:25
Solution:
As y and n are both integers so y is of the form of 3*5*2*2*K
where K= an integer.
to make n as an integer and 450y to be a cube of a number.
Now checking with the answers we get only 1 as K.
So the answer is (1).
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Re: 450y=n^3, finding integer  [#permalink]

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New post 09 Mar 2010, 04:27
I take (1) as solution even though (5) also satisfies as GMAT says to go with the first right answer.
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Re: 450y=n^3, finding integer  [#permalink]

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New post 09 Mar 2010, 13:26
2
450 * y = n*n*n
Prime factorization of 450 yields 450 = 5*5*2*3*3

5*5*2*3*3 * (y) = (5*2*3...)*(5*2*3...)*(5*2*3...)
==> y needs to contain at least 5,2,2,3 ==> 5, 2^2,3

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Re: 450y=n^3, finding integer  [#permalink]

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New post 09 Mar 2010, 15:37
mustdoit wrote:
If n and y are positive integers and 450y = n^3, which of the following must be an integer?

1 - y/(3*2^2*5)
2 - y/(3^2*2*5)
3 - y/(3*2*5^2)
4 - y/(5^3*2^2)
5 - y/(5*3*2)


What is the source of this question? I don't like this question!! GMAT does not have two correct answer options. Stop practising whatever prep material that is. Wrong materials are lethal to scores!!!!!
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Re: 450y=n^3, finding integer  [#permalink]

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New post 09 Mar 2010, 21:01
1
mustdoit wrote:
If n and y are positive integers and 450y = n^3, which of the following must be an integer?

1 - y/(3*2^2*5)
2 - y/(3^2*2*5)
3 - y/(3*2*5^2)
4 - y/(5^3*2^2)
5 - y/(5*3*2)



Given 450y = n^3

this gives y= (n^3)/450

Now the prime factorisation of 450 gives us = (5^2)*(3^2)*(2)

Now we have y= n^3/ (5^2)*(3^2)*(2)

we know that y is an integer therefore n would have each of these three that is 5,3 and 2 as its prime factors.

you have one 2, two 3s and two 5s in the denominator. What prime factors COULD n have that would make this relatively close? 2, 3 and 5 naturally!


looking at the factors...

(2,3,3,5,5) * (?) = (2,3,5) * (2,3,5) * (2,3,5) = (2,2,2,3,3,3,5,5,5)

therefore the prime factors of y must be AT MINIMUM

(2,2,3,5) hence the anwer is only A and not E.

Hope this helps !!!
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Re: 450y=n^3, finding integer  [#permalink]

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New post 09 Mar 2010, 21:09
mustdoit wrote:
If n and y are positive integers and 450y = n^3, which of the following must be an integer?

1 - y/(3*2^2*5)
2 - y/(3^2*2*5)
3 - y/(3*2*5^2)
4 - y/(5^3*2^2)
5 - y/(5*3*2)



Errata - the last option is y/(5^3*2)
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Re: 450y=n^3, finding integer  [#permalink]

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New post 26 Aug 2012, 09:45
mustdoit wrote:
mustdoit wrote:
If n and y are positive integers and 450y = n^3, which of the following must be an integer?

1 - y/(3*2^2*5)
2 - y/(3^2*2*5)
3 - y/(3*2*5^2)
4 - y/(5^3*2^2)
5 - y/(5*3*2)



Given 450y = n^3

this gives y= (n^3)/450

Now the prime factorisation of 450 gives us = (5^2)*(3^2)*(2)

Now we have y= n^3/ (5^2)*(3^2)*(2)

we know that y is an integer therefore n would have each of these three that is 5,3 and 2 as its prime factors.

you have one 2, two 3s and two 5s in the denominator. What prime factors COULD n have that would make this relatively close? 2, 3 and 5 naturally!


looking at the factors...

(2,3,3,5,5) * (?) = (2,3,5) * (2,3,5) * (2,3,5) = (2,2,2,3,3,3,5,5,5)

therefore the prime factors of y must be AT MINIMUM

(2,2,3,5) hence the anwer is only A and not E.

Hope this helps !!!


Mustdoit Your calculation are right but you missed that if y/(3.5.2^2) is an integer then y/(3.5.2) will also be an integer at any cost. So this far both options A & E are right.
Now as you have updated the choice E, the answer will be A
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Re: 450y=n^3, finding integer  [#permalink]

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New post 26 Aug 2012, 14:06
A and E. Both are correct.

I think there is a typo in E.
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Re: If n and y are positive integers and 450y = n^3, which of  [#permalink]

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New post 26 Aug 2012, 21:35
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Re: If n and y are positive integers and 450y = n^3, which of  [#permalink]

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Re: If n and y are positive integers and 450y = n^3, which of &nbs [#permalink] 14 Jul 2018, 11:40
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