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by mike22629 » Sat May 09, 2009 9:02 am
If n and y are positivie integers and 450y = n^3, then 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))

A. None
B. I only
C. II only
D. III only
E. I,II, and III







I assume it has something to do with finding the factors of 450, but can not find out a quick way to find a multiple of 450 that is a cube.

I will be back in a few hours. I do not want to post OA yet, to see responses, but do not worry, will post in a few hours.
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Re: GMAT Prep

by Stuart@KaplanGMAT » Sat May 09, 2009 9:52 am
mike22629 wrote:If n and y are positivie integers and 450y = n^3, then 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))

A. None
B. I only
C. II only
D. III only
E. I,II, and III

I assume it has something to do with finding the factors of 450, but can not find out a quick way to find a multiple of 450 that is a cube.
When in doubt (on factoring questions), break things down into primes!

450 = 3*150 = 3*3*50 = 3*3*5*10 = 3*3*5*5*2

Now, we know that 450y is a perfect cube. Just as perfect squares have pairs of prime factors, perfect cubes have triplets of prime factors.

Therefore the smallest possible value of y "fills out" the triplets started by 450.

We have one "2", so we need two more.
We have two "3"s, so we need one more.
We have two "5"s, so we need one more.

Accordingly, the smallest value of y is 2*2*3*5.

Plugging into our roman numerals:

I) 2*2*3*5/2*2*3*5 IS an integer: eliminate (A), (C) and (D).

II) 2*2*3*5/2*3*3*5 is NOT an integer: eliminate (E).

Only (B) remains, no need to even check III.
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by mike22629 » Sat May 09, 2009 12:26 pm
Perfect explanation Stuart. Thanks a lot!
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