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If n and y are positive integers and 450y = n^3 which of the following

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abhi758
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If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   Updated on: 07 Oct 2019, 02:58
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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
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B

Originally posted by abhi758 on 11 Apr 2010, 23:56.
Last edited by Bunuel on 07 Oct 2019, 02:58, edited 2 times in total.
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   12 Apr 2010, 05:54
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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

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

\(450y=n^3\) --> \(2*3^2*5^2*y=n^3\). As \(n\) 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 \(2*3^2*5^2*y=(2*3*5)^3=n^3\). 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 which of the following [#permalink] 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.

Greets
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink] 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 which of the following [#permalink] 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: If n and y are positive integers and 450y = n^3 which of the following [#permalink] 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: If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   12 Oct 2010, 00:14
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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: If n and y are positive integers and 450y = n^3 which of the following [#permalink] 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 which of the following [#permalink] New post   22 May 2014, 13:11
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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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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   22 May 2014, 20:25
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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 which of the following [#permalink] New post   23 May 2014, 10:53
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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.


Similar questions to practice:
if-n-is-a-positive-integer-and-n-2-is-divisible-by-96-then-127364.html
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-90523.html
a-certain-clock-marks-every-hour-by-striking-a-number-of-tim-91750.html
if-m-and-n-are-positive-integer-and-1800m-n3-what-is-108985.html
if-x-and-y-are-positive-integers-and-180x-y-100413.html
n-is-a-positive-integer-and-k-is-the-product-of-all-integer-104272.html
if-x-is-a-positive-integer-and-x-2-is-divisible-by-32-then-88388.html
if-n-and-y-are-positive-integers-and-450y-n-92562.html
if-5400mn-k-4-where-m-n-and-k-are-positive-integers-109284.html
if-n-and-y-are-positive-integers-and-450y-n-92562.html

Hope this helps.
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   30 Jun 2017, 00:04
Cant we take higher values of N and Y so that it satisfies the equation,the all 3 possiblities are correct and answer is E
eg:
y= 5^4*3^4*2^5
N=5^2*3^2*2^2
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   30 Jun 2017, 00:18
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jakexix wrote:
Cant we take higher values of N and Y so that it satisfies the equation,the all 3 possiblities are correct and answer is E
eg:
y= 5^4*3^4*2^5
N=5^2*3^2*2^2


The question asks which of the following MUST be an integer, not COULD be an integer. If y = 2^2*3*5, which is the least possible value of y, then only B is true. Please re-read the thread and follow the links to similar questions to understand the concept better.
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   24 Jan 2020, 22:50
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abhi758 wrote:
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


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

For a number to be perfect cube the powers of all its prime factors must be multiples of 3

i.e. We need minimum \(2^2*3*5\) in y in order to make the number on left hand side a perfect cube

i.e. y must be a multiple of \(2^2*3*5\) but we can't be certain of any higher powers of respective prime factors than what is minimum hence

Answer: Option B
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   09 Oct 2022, 04:29
rajatjain14 wrote:
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


Can you please help me out that how y=2^2*3*5 and how it halps to make 450*y a perfect cube

I am a toddler in these concepts
Thanks

Posted from my mobile device
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   10 Oct 2022, 16:01
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abhi758 wrote:
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



Here is what I think:

Given that 450*y = n^3 -------------- y = n^3 / 450
450 by prime factorization = 2 * 3^2 * 5^2

Since y = integer and y > 0 then n^3 / 450 should give an integer -------- which means that n^3 should be completely divisible by 2 * 3^2 * 5^2 (=450)

This follows that the min value of n will be 2 * 3 * 5 ----------- meaning n^3 will be 2^3 * 3^3 * 5^3

Y = n^3 / 450 = (2^3 * 3^3 * 5^3) / 2 * 3^2 * 5^2
Y = 2^2 * 3 * 5

Now if you look at the options only option (i) will work.

Answer – B
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   15 Oct 2022, 08:00
varshas044 wrote:
abhi758 wrote:
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



Here is what I think:

Given that 450*y = n^3 -------------- y = n^3 / 450
450 by prime factorization = 2 * 3^2 * 5^2

Since y = integer and y > 0 then n^3 / 450 should give an integer -------- which means that n^3 should be completely divisible by 2 * 3^2 * 5^2 (=450)

This follows that the min value of n will be 2 * 3 * 5 ----------- meaning n^3 will be 2^3 * 3^3 * 5^3

Y = n^3 / 450 = (2^3 * 3^3 * 5^3) / 2 * 3^2 * 5^2
Y = 2^2 * 3 * 5

Now if you look at the options only option (i) will work.

Answer – B



Greatly explained. Huge thanks!
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink] New post   29 May 2023, 06:50
For such problems, I've found it useful to remember that the only way we can get an integer from a given unknown fraction is if the denominator divides into the numerator fully. In other words, the numerator is REQUIRED to contain the ENTIRE prime factorization of the denominator.

Once this condition is met, the numerator is free to additional copies of the existing prime factors or even other new factors - more the merrier as it only increases the numerator's divisibility.
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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