Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

"Must be an integer" means for the lowest possible value of \(y\).

\(450y=x^3\) --> \(2*3^2*5^2*y=x^3\). As \(x\) and \(y\) are integers, \(y\) must complete the powers of 2, 3, and 5 to cubes (generally to the multiple of 3). Thus \(y_{min}=2^2*3*5\), in this case \(2*3^2*5^2*y=(2*3*5)^3=x^3\). Notice that for this value of \(y\) only the first option is an integer: \(\frac{y}{{3*2^2*5}}=\frac{2^2*3*5}{{3*2^2*5}}=1\).

Re: Number Properties [#permalink]
12 Apr 2010, 22:15

Moderator,

Can you elaborate more on the solution, seems unfathomable for simple minds like mine .

How did we deduce that "Minimum value of Y"? _________________

------------------------------------------------------------------------- Ros. Nice Post + Some help + Lucid solution = Kudos

The greatest pleasure in life is doing what people say you cannot do | Great minds discuss ideas, average minds discuss events, small minds discuss people. -------------------------------------------------------------------------

Re: Number Properties [#permalink]
13 Apr 2010, 01:58

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

shrivastavarohit wrote:

Moderator,

Can you elaborate more on the solution, seems unfathomable for simple minds like mine .

How did we deduce that "Minimum value of Y"?

\(x\) and \(y\) are integers and \(450y=x^3\) --> \(450y\) equals to cube of an integer. \(450y=2*3^2*5^2*y=x^3\). The smallest value of \(y\) for which \(2*3^2*5^2*y\) is a cube of an integer is when \(y=2^2*3*5\). In this case \(450y=(2*3^2*5^2)*(2^2*3*5)=(2*3*5)^3\). Of course \(y\) can take another values as well, for example \(y=2^5*3^4*5^7\) and in this case \(450y=(2*3^2*5^2)*(2^5*3^4*5^7)=(2^3*3^2*5^3)^3\), but the smallest value of \(y\) is when \(y=2^2*3*5\).

Re: Number Properties [#permalink]
13 Apr 2010, 02:28

Thanks for the quick response it makes more sense now I will check out the problems (similar) on the threads mentioned by you.

Is this really a GMAT question from the calculation it looks solving this should take more than 2 mins.

Posted from my mobile device _________________

------------------------------------------------------------------------- Ros. Nice Post + Some help + Lucid solution = Kudos

The greatest pleasure in life is doing what people say you cannot do | Great minds discuss ideas, average minds discuss events, small minds discuss people. -------------------------------------------------------------------------

Re: Number Properties [#permalink]
13 Apr 2010, 02:32

Oops !!! One more important thing why is plugging in values applicable to these kind of problems?

It seemed pretty simple from the question to plug in 2 integer values in the final equation and check for the integer output btw which miserably failed on all three equations.

Posted from my mobile device _________________

------------------------------------------------------------------------- Ros. Nice Post + Some help + Lucid solution = Kudos

The greatest pleasure in life is doing what people say you cannot do | Great minds discuss ideas, average minds discuss events, small minds discuss people. -------------------------------------------------------------------------

Re: problem solving [#permalink]
27 Jun 2010, 05:37

1

This post received KUDOS

I am not sure if I got the question right. The way I understand this Q isthat 450y = n*n*n and the question is whether the given options are integers.

The answer of above Q is as follows: We know that: n = cubic root (450y) = cubic root (2X3X3X5X5) For n to be an integer, y should be factor of 2X2X3X5.

So y=(2X2X3X5)k where k is any natural number. if k = 1, y=2X2X3X5 so the answer is b.

I hope this explanation meets your satisfaction.

Last edited by jakolik on 27 Jun 2010, 20:35, edited 1 time in total.

Re: Tough integer properties question! [#permalink]
11 Oct 2010, 23:14

1

This post received KUDOS

atomjuggler wrote:

Raising the white flag on this one. I really hope I'm not missing something obvious Source: GMATPrep Test #1

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

I. \(\frac{y}{3*2^2*5}\)

II. \(\frac{y}{3^2*2*5}\)

III. \(\frac{y}{3*2*5^2}\)

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

\(450*y=n^3\) Since 450y is a whole cube, in the prime factorization of 450y, all the exponents on the prime factors must be multiples of 3. \(450*y = y * 3^2 * 5^2 * 2\) So y must have atleast \(3, 5, 2^2\) as prime factors. y must be of the form form 3x5x4xA where A is also another perfect cube. By this logic the answer is (I) only or B, as in all cases y/(3x5x4) has to be an integer. (ii) or (iii) cannot be true if A=1

Re: 450y=n^3, finding integer [#permalink]
19 Jun 2012, 21:45

2

This post received KUDOS

Expert's post

enigma123 wrote:

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

1. \(\frac{Y}{3 * 2^2 * 5}\)

2. \(\frac{Y}{3^2 * 2 * 5}\)

3. \(\frac{Y}{3 * 2 * 5^2}\)

A) None

B) I only

C) II Only

D) III only

E) I, II and III onmly.

The critical point in this question is that n and y are integers. Since n is an integer, n^3 must be the cube of an integer. So in n^3, all the prime factors of n must be cubed (or have higher powers which are multiples of 3)

Now consider 450y = n^3 450 is not a perfect cube. So whatever is missing in 450, must be provided by y to make a perfect cube.

\(450 = 45*10 = 2 * 3^2 * 5^2\)

To make a perfect cube, we need at least two 2s, a 3 and a 5. These missing factors must be provided by y. Therefore, \(2^2*3*5\) must be a factor of y. This means \(y/2^2*3*5\) must be an integer. _________________

Re: If n and y are positive integers and 450y=n^3 [#permalink]
28 Oct 2013, 12:36

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If n and y are positive integers and 450y=n^3 [#permalink]
19 Apr 2014, 21:11

Concept Tested:

For a perfect square the powers should occur in pairs.

For a perfect cube powers should occur in cubes.

450y = x^3

5^2 * 3^2 * 2 * y = x^3

For LHS to become a cube Y must be

y= 5*3*2^2

Only (1) matches it. _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: If n and y are positive integers and 450y=n^3 [#permalink]
22 May 2014, 12:11

Hi rrsnathan,

Your approach is wrong because you are assuming 1 fixed value of \(x = 2*3^2*5^2\)

But question explicitly asks which option MUST be an integer always. In your assumption, Option C comes out to be an integer but that may not be true in other cases.

So, Right method to solve it is:

\(450y = x^3\)

\((2*3^2*5^2) * y = x^3\)

For LHS to be a cube (As R.H.S is a cube),

\(y = 2^2*3*5*k^3\) (where k is a positive integer)

Hence, Option (b) is always correct

Rgds, Rajat _________________

If you liked the post, please press the'Kudos' button on the left

Re: If n and y are positive integers and 450y=n^3 [#permalink]
22 May 2014, 19:25

Expert's post

rrsnathan wrote:

I could understand the options

But tell y my alternate approach is wrong 450y = x^3

X and Y are integer 450 = 2*3^2*5^2

X=((2*3^2*5^2)y)/x^2 Note: as x is an integer if x = 2*3^2*5^2

X = y/ (2*3^2*5^2)

This is incorrect. Note that in the denominator, you have x^2, not x. You are assuming \(x = 2*3^2*5^2\) and putting it as it is in the denominator. Also note that since you have the same x on the left hand side, if you put \(x = 2*3^2*5^2\), you get

\(2*3^2*5^2 = (2*3^2*5^2)*y/(2*3^2*5^2)^2\) i.e. \(y = (2*3^2*5^2)^2\)

Basically, you have assumed a value of x and got a value of y. There is no reason for you to assume the value of x as \(2*3^2*5^2\).

The point here is that y must have some values such that when it multiplies 450, it gives a perfect cube. This is the method discussed above in my post. _________________

HBS: Reimagining Capitalism: Business and Big Problems : Growing income inequality, poor or declining educational systems, unequal access to affordable health care and the fear of continuing economic distress...

I am not panicking. Nope, Not at all. But I am beginning to wonder what I was thinking when I decided to work full-time and plan my cross-continent relocation...

Over the last week my Facebook wall has been flooded with most positive, almost euphoric emotions: “End of a fantastic school year”, “What a life-changing year it’s been”, “My...