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If n and y are positive integers and 450y=n^3

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

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New post Updated on: 12 Apr 2010, 00:58
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If x and y are positive integers and 450y=x^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

Please explain your answers..


Originally posted by abhi758 on 11 Apr 2010, 23:56.
Last edited by abhi758 on 12 Apr 2010, 00:58, edited 1 time in total.
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Re: Number Properties  [#permalink]

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New post 12 Apr 2010, 05:54
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abhi758 wrote:
If x and y are positive integers and \(450y=x^3\), 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

Please explain your answers..



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

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

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New post 31 Oct 2013, 05:49
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I finally get it. If you want the cuberoot of 450y to be an integer, you need the primefactors to be cubed.

So the primefactors we have for 450 are 2, 3², 5². To be perfect cubes, we need to multiply by 2², 3, 5 which in this case equals y.

so we get that (2*3²*5²) * (2²*3*5) = 2³*3³*5³. If we'd cuberoot x now, we'd get an integer.

This leads to the conclusion that y / 2²*3*5 = 1 = integer. So answer B.

I tried explaining in simpler words so I'd get it myself, I hope it helps the you other non-math-overbrains.

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Re: Number Properties  [#permalink]

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New post 12 Apr 2010, 23:15
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Moderator,

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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Re: Number Properties  [#permalink]

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New post 13 Apr 2010, 02:58
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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 \(y=2^2*3*5\).

You can check the similar problems at:
og-quantitative-91750.html#p704028
division-factor-88388.html#p666722
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Re: Number Properties  [#permalink]

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New post 13 Apr 2010, 03:28
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.

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Re: Number Properties  [#permalink]

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New post 13 Apr 2010, 03:32
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.

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Re: Number Properties  [#permalink]

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New post 14 Apr 2010, 23:00
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Thanks Bunnel! your first explanation makes it crystal clear..
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Re: problem solving  [#permalink]

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New post Updated on: 27 Jun 2010, 21:35
1
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.

Originally posted by jakolik on 27 Jun 2010, 06:37.
Last edited by jakolik on 27 Jun 2010, 21:35, edited 1 time in total.
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Re: Tough integer properties question!  [#permalink]

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New post 12 Oct 2010, 00:14
1
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

Answer is (b)
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Re: Algebra  [#permalink]

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New post 08 Jan 2012, 05:29
kotela wrote:
Can anyone please help me in solving this problem.......


lets find prime factors of 450 = 2* 3* 3* 5* 5

for 450* y to be equal to n^3 450y needs to have 2,3and 5 three times ( as n is an integer )

i.e. 450 y = (2* 3* 3* 5* 5) (2*2*3*5) * x = n^3

now y = (2*2*3*5) * x

only option 1 will then give an integer
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Re: 450y=n^3, finding integer  [#permalink]

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New post 19 Jun 2012, 22:45
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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.
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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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New post 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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New post 22 May 2014, 11:59
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)

And its option c).

Please correct me.

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

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New post 22 May 2014, 13:11
1
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

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

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New post 22 May 2014, 20:25
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.
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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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New post 23 May 2014, 10:53
1
3
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)

And its option c).

Please correct me.

Thanks in advance.


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

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New post 16 Jul 2015, 09:39
450 = (5^2) * (3^2) * (2)

so, in order for 450y to be a cube of some number, y must be equal to 5*3*(2^2) = 60.

Only option (i) gives an integer (60/60). So, Ans(B).
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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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New post 29 Nov 2016, 20:13
Our goal is to get figure out why so that x is a perfect square:

450 = (2)(3^2)(5^2)

To make a perfect square we are missing the following: (2^2)(3)(5) --> This is the minimum value y must take on in order for us to get the cube root of x to be an integer.

Option I is the only choice given that will give us an integer value for x. Thus, B is the correct answer.
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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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New post 28 Dec 2016, 11:08
Great Official Question.
Here is what i did in this one =>

x is an integer => x^3 must be a perfect cube.

Hence the general expression for y => 5*3*2^2*prime^3

Hence least value of y will be 5*3*2^2

Only 1 is in integer

Hence B

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Re: If n and y are positive integers and 450y=n^3 &nbs [#permalink] 28 Dec 2016, 11:08

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