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GMAT Prpe

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by moneyman » Mon Mar 31, 2008 2:39 am
I dont know what is the component I should be focussing on while doing this problem

If n and y 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

Thanks
Maxx
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Re: GMAT Prpe

by camitava » Mon Mar 31, 2008 2:45 am
moneyman wrote:I dont know what is the component I should be focussing on while doing this problem

If n and y 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

Thanks
450 = 2 x 5^2 x 3^2
Now 2 x 5^2 x 3^2 x y = n^3
So at minimum, y = 2^2 x 5 x 3
So y/3*2^2*5 is the minimum value which will always be integer. Got me, Max?
Correct me If I am wrong


Regards,

Amitava
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by moneyman » Mon Mar 31, 2008 9:09 am
Got it Amitava..But I really could not figure out the approach..now its clear..Thanks
Maxx
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folks

by resilient » Mon Mar 31, 2008 8:07 pm
i tried to push numbers out for this and wound up with n=10 and y=2.2

when i plugged them into the answer choices not one of the roman numerals gave an integer. So my first answer is none will give an integer.


I tried to break this up into primes but then got lost.

can someone explain this to me in baby steps and please give qa. thanks folks!
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Re: folks

by tmmyc » Mon Mar 31, 2008 9:10 pm
Enginpasa1 wrote:i tried to push numbers out for this and wound up with n=10 and y=2.2

when i plugged them into the answer choices not one of the roman numerals gave an integer. So my first answer is none will give an integer.


I tried to break this up into primes but then got lost.

can someone explain this to me in baby steps and please give qa. thanks folks!
Step by step:

Find the prime factors of 450.
450
->45*10
->9*5*2*5
-->3*3*5*2*5
--->2*3*3*5*5

The question states that 450y = n^3.
Since y and n are both positive integers, this means that 450y must be a perfect cube.

In order for 450y to be a perfect cube, we look at our current prime factors of 450:
(2) (3*3) (5*5)

Therefore, in order for 450y to be a perfect cube, we need an additional 5, an additional 3, and two additional 2's.

(2) (3*3) (5*5) * y
->(2) (3*3) (5*5) * [(2*2) (3) (5)]
-->(2*2*2) (3*3*3) (5*5*5) = perfect cube

Thus, at minimum, y must be [(2*2) (3) (5)] or 2^2 * 3 * 5.

We look at our answer choices and plug in for y.

I)
y/3*2^2*5
-> (2^2 * 3 * 5) / 3*2^2*5 = 1 (integer)

II)
y/3^2*2*5
-> (2^2 * 3 * 5) / 3^2*2*5 = 2/3 (not an integer)

III)
y/3*2*5^2
-> (2^2 * 3 * 5) / 3*2*5^2 = 2/5 (not an integer)
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this is different

by resilient » Mon Mar 31, 2008 11:57 pm
Is there a way to approach this problem and not go through all of your steps. Can we force a value for y and n and then test each roman numeral?
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Re: this is different

by tmmyc » Tue Apr 01, 2008 10:24 am
Enginpasa1 wrote:Is there a way to approach this problem and not go through all of your steps. Can we force a value for y and n and then test each roman numeral?
To me, this approach makes the most sense, and like camitava showed, it is pretty straightforward. I would not go through my step by step method on paper; rather, a lot of it would be mental.

The underlying process is quite simple:
1. Recognize that 450y must be a perfect cube.
2. Break 450 into its prime factors and see what y at minimum must be to create a perfect cube.
3. Glance at I, II, and III to see if this minimum value of y can cancel all the denominator values.
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hinging

by resilient » Tue Apr 01, 2008 11:16 am
i guess the answer hinges on seeing that 450y must be a perfect cube. NOw I get it! thanks
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by humeixia » Fri Apr 11, 2008 3:53 pm
Great solution. :D
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Re: folks

by Musiq » Fri Apr 11, 2008 9:57 pm
Enginpasa1 wrote:i tried to push numbers out for this and wound up with n=10 and y=2.2

when i plugged them into the answer choices not one of the roman numerals gave an integer. So my first answer is none will give an integer.


I tried to break this up into primes but then got lost.

can someone explain this to me in baby steps and please give qa. thanks folks!
A quick suggestion Enginpasa:

On the GMAT, more often than not, there is as much information in the options and the other parts of the question as the actual question itself.

The 3 Roman Numeral statements are CLEAR indicators that Prime factorisation is in some way or the other involved.

Any method that takes us away from that will be longer and relatively more cumbersome.
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by Matmasi » Wed Sep 16, 2009 2:32 pm
Thank you very much
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Re: folks

by Matmasi » Wed Sep 16, 2009 2:34 pm
tmmyc wrote:
Enginpasa1 wrote:i tried to push numbers out for this and wound up with n=10 and y=2.2

when i plugged them into the answer choices not one of the roman numerals gave an integer. So my first answer is none will give an integer.


I tried to break this up into primes but then got lost.

can someone explain this to me in baby steps and please give qa. thanks folks!
Step by step:

Find the prime factors of 450.
450
->45*10
->9*5*2*5
-->3*3*5*2*5
--->2*3*3*5*5

The question states that 450y = n^3.
Since y and n are both positive integers, this means that 450y must be a perfect cube.

In order for 450y to be a perfect cube, we look at our current prime factors of 450:
(2) (3*3) (5*5)

Therefore, in order for 450y to be a perfect cube, we need an additional 5, an additional 3, and two additional 2's.

(2) (3*3) (5*5) * y
->(2) (3*3) (5*5) * [(2*2) (3) (5)]
-->(2*2*2) (3*3*3) (5*5*5) = perfect cube

Thus, at minimum, y must be [(2*2) (3) (5)] or 2^2 * 3 * 5.

We look at our answer choices and plug in for y.

I)
y/3*2^2*5
-> (2^2 * 3 * 5) / 3*2^2*5 = 1 (integer)

II)
y/3^2*2*5
-> (2^2 * 3 * 5) / 3^2*2*5 = 2/3 (not an integer)

III)
y/3*2*5^2
-> (2^2 * 3 * 5) / 3*2*5^2 = 2/5 (not an integer)
thank you very much, I was surfing the web looking for such a good explanation. Good luck
mat
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