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A question that I'm sure is easy that I screwed up.

Expert replies
by Kobe_Kassidy » Tue Aug 07, 2012 9:04 am
Hey guys. Got myself stumped on this one.

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

The correct answer for this question ended up being: y/(3 X 2^2 X 5) (three x two squared x five)

How do you get to this solution? There has to be a rule or method I am not recognizing.
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Source: — Problem Solving |

by GMATGuruNY » Tue Aug 07, 2012 11:25 am
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by Kobe_Kassidy » Tue Aug 07, 2012 12:30 pm
Thank you for your help!

In studying the principles and subjects for the GMAT, where would I come across the subject tested in this problem? Perfect cubes are certainly not new to me, but admittedly, I didn't know that every perfect cube had to have 3 of every prime factor.

Also, is there a search function where I can easily find if questions I am asking have been asked before? I am just going through the free GMAT software practice exams.

Thank you!
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by GMATGuruNY » Wed Aug 08, 2012 7:13 am
Kobe_Kassidy wrote:Thank you for your help!

In studying the principles and subjects for the GMAT, where would I come across the subject tested in this problem? Perfect cubes are certainly not new to me, but admittedly, I didn't know that every perfect cube had to have 3 of every prime factor.
This problem is about PRIME-FACTORIZATION and DIVISIBILITY.

450y/n³ = 1
2*3*3*5*5*y / n*n*n = 1.

Since there are factors of 2, 3 and 5 in the numerator, there must be factors of 2, 3, and 5 in the denominator.
Thus, the smallest possible value of n = 2*3*5, yielding the following:
2*3*3*5*5*y / (2*3*5) (2*3*5) (2*3*5) = 1.

Since there must be the same number of 2's, 3's, and 5's in the numerator and there are in the denominator, the smallest possible value of y = 2*2*3*5.
Also, is there a search function where I can easily find if questions I am asking have been asked before? I am just going through the free GMAT software practice exams.

Thank you!
In the upper right corner is a SEARCH BAR.
To search for a problem, just type into the search bar a few keywords from the problem.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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by Brent@GMATPrepNow » Wed Aug 08, 2012 9:14 am
Kobe_Kassidy wrote:Hey guys. Got myself stumped on this one.

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

The correct answer for this question ended up being: y/(3 X 2^2 X 5) (three x two squared x five)

How do you get to this solution? There has to be a rule or method I am not recognizing.
450y=n^3
2*3*3*5*5*y = n^3
For 2*3*3*5*5*y to be a cube, we need the number of 2's, 3's and 5's to be divisible by 3.
So, for example, 2*2*2*3*3*3*5*5*5 = (2*3*5)^3
For 2*3*3*5*5*y to be a cube, it must be the case that y has at least 2 additional 2's, 1 additional 3 and 1 additional 5.
So, y = 2*2*3*5*(other possible numbers)
We see that 2*2*3*5 is the smallest possible value for y.

Of the possible answers in the original question, only the first one (y/ 3*2^2*5) will simplify to be an integer.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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