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

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

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New post 29 Dec 2016, 12:10
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..



\(450y=x^3\)

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

The minimum value of \(y =\) \(2^2*5*3\)

Hence, among the given options, only (I) is possible...

So, correct answer must be (B)

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

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New post 22 Jan 2017, 07:58
1)If cube root of 450y is an integer, then 450y has to have three of each prime factors. 2)Prime factors of 450 are 2*5*5*3*3. It is missing two 2's, one 5, and one 3. 3)To compensate for these missing factors y has to have all of them as its prime factors, consequently y has to be divisible by 2^2*5*3.
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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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New post 26 Jan 2017, 16:24
1) x is cubed, which means that its prime factors can be divided into three identical sets.
2) 450*y should have the same three sets of prime factors as x cubed, since the two expressions are equal
3) The prime factors of 450 are 2*5*5*3*3. This means that y has to consist of 2^2*5*3 at least, and it has to be divisible by this expression.

The correct answer is B.
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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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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  [#permalink]

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New post 30 Jun 2017, 00:18
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  [#permalink]

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New post 03 Dec 2018, 06:19
Bunuel wrote:
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.



Hey Bunuel,

I have a query. Why are we considering only minimum value of y?

y = 2^8 * 3^7 * 5^7 also gives a integer value for n.
In that case, all the 3 options gives integers.

Please clarify. What am I missing here?
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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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New post 09 Apr 2019, 11:44
loneRanger1050 wrote:
Bunuel wrote:
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.



Hey Bunuel,

I have a query. Why are we considering only minimum value of y?

y = 2^8 * 3^7 * 5^7 also gives a integer value for n.
In that case, all the 3 options gives integers.

Please clarify. What am I missing here?


Hi loneRanger1050,

I understand as to what led you to think that all the three choices could suffice .


Let's see what led you to this conclusion.

if write n^3 in its prime factor form
\(n^{3}\)= \(5^{2}\)\(3^{2}\) 2 y

Now since n is a integer , say n=k, then \(n^{3}\)= \(k^{3}\), which means if we factorize n in its prime factors, then for \(n^{3}\) the prime factorization should have powers which are multiples of 3 .

Then y can be of the form =\(2^{2}\)\(2^{P}\) \(3^{1}\)\(3^{Q}\)\(5^{1}\)\(5^{R}\) \(A^{Z}\)
where P, Q R Z must be multiple of 3 and A is a prime number ( other than 2,3,5)

But notice the question Stem. It asks for a scenario in which \(MUST BE\) true. Had they asked for a scenario of CAN BE TRUE your logic would be correct.

Since we are asked for must be true scenario, then the condition must also exist for smallest possible value of y also . And least value of y can be \(2^{2}\)\(3^{1}\)\(5^{1}\)

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 15 Apr 2019, 08:23
For those of you, who are more comfortable plugging in values, follow the below approach:-

450*Y=X^3

Now, simply think of numbers that are perfect cubes

we have 8,27,64 etc.

Now, try to get 2700, so we have

450*60=30^3

Perfect! Now Y=60, X=30

Test the answers now:-
1. We have 60/60 - an integer
2. 60/90-not an integer
3. 60/150-not an integer.
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Re: If n and y are positive integers and 450y=n^3   [#permalink] 15 Apr 2019, 08:23

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