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"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: \(\frac{y}{{3*2^2*5}}=\frac{2^2*3*5}{{3*2^2*5}}=1\).

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

How did we deduce that "Minimum value of Y"?
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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 \(y=2^2*3*5\).

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.

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------------------------------------------------------------------------- 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. -------------------------------------------------------------------------

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.

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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, 21:35, edited 1 time in total.

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

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.
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Re: If n and y are positive integers and 450y=n^3 [#permalink]

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28 Oct 2013, 13:36

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Re: If n and y are positive integers and 450y=n^3 [#permalink]

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19 Apr 2014, 22:11

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.
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Re: If n and y are positive integers and 450y=n^3 [#permalink]

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22 May 2014, 13:11

1

This post received KUDOS

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
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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.
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Re: If n and y are positive integers and 450y=n^3 [#permalink]

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16 Jul 2015, 07:55

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).

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