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What is the smallest positive integer n such that 6,480n^1/2

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What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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New post 19 Mar 2014, 06:23
SOLUTION

What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

First make prime factorization of 6,480: \(6,480=2^4*3^4*5\).

Next, for \(2^4*3^4*5*\sqrt{n}\) to be a perfect cube \(\sqrt{n}\) must complete the powers of 2, 3 and 5 to a multiple of 3, thus the least value of \(\sqrt{n}\) must equal to \(2^2*3^2*5^2=900\). Thus the least value of \(n\) is \(900^2=30^4\).

Answer: E.
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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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New post 19 Mar 2014, 07:15
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Bunuel wrote:
What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4


Sol: Let's factorize 6480 and we get 6480= 3^4*2^4*5
Now we need to see for what minimum value of \(\sqrt{n}\)*6480= a^3 where a is an Integer

So from 6480 we already have 2^4*3^4*5*\(\sqrt{n}\) = (2^2)^3* (3^2)^3*(5)^3 why cause a is an integer which will need to be have the same factors which are in LHS

solving for \(\sqrt{n}\) = (2^6*3^6*5^3)/ 2^4*3^4*5 and we get

\(\sqrt{n}\)= 2^2*3^2*5^2
Or n = 2^4*3^4*5^4 or 30^4

Ans is E
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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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New post 19 Mar 2014, 19:13
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Answer = E. \(30^4\)

Looking at the options, we see that 5 & 30 are not perfect square roots; so option A & C can be discarded

\(6480 = 2^3 . 3^4 . 10\)

\(2^3\) is already a perfect cube; no further adjustment is required.

\(3^4\) requires \(3^2\) to make it perfect cube

10 requires \(10^2\) to make it perfect cube

So we require \(30^2\) to make the term a perfect cube

\(\sqrt{n} = 30^2\)

\(n = 30^4\) = Answer
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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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New post 19 Mar 2014, 20:59
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6480 = 8*810 = 8*9*90 = 8*27*3*5*2
8 and 27 are perfect cube
To make a perfect cube, we need \(3^2\)*\(5^2\)*\(2^2\)
hence\(\sqrt{n} = 3^2*5^2*2^2\)
therefore \(n^2\) = \(3^4\)*\(5^4\)*\(2^4\) = \((3*5*2)^4\) = \(30^4\)

Answer - E
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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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SOLUTION

What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

First make prime factorization of 6,480: \(6,480=2^4*3^4*5\).

Next, for \(2^4*3^4*5*\sqrt{n}\) to be a perfect cube \(\sqrt{n}\) must complete the powers of 2, 3 and 5 to a multiple of 3, thus the least value of \(\sqrt{n}\) must equal to \(2^2*3^2*5^2=900\). Thus the least value of \(n\) is \(900^2=30^4\).

Answer: E.
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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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New post 19 Dec 2015, 10:53
for 6480* sqrt(n) to be perfect cube, we need to find first the prime factorization of 6480.
well, 6480 = 648*10.
648-> sum of the digits is divisible by 9, so let's divide by 9.
648=9*72=>9*9*8.
ok, so 6480=(2^4)*5*(3^4).
for the number 6480* sqrt(n) to be perfect square, sqrt(n) must be at least 2^2*5^2*3^2. since it is sqrt, then n must be 2^4*5^4*3^4 or (2*3*5)^4 or 30^4, which is E.
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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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New post 03 Dec 2016, 20:02
Great Official Question.
Here is what i did->
Let √n=p
now 6480 = 2^4*5*3^4
the minimum value of p must be 2^2*5^2*3^2 for the product to be a perfect cube.
Hence √n=30^2
so n=30^4
hence E

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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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New post 25 Dec 2017, 10:32
Bunuel wrote:
What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4


Prime Factorice 6480
=> \(6480=(2^4*3^4*5)\)
=> Required \(6480*\sqrt{n}=perfect cube=(2^4*3^4*5)*\sqrt{n}\)
=> This set of prime factors \((2^4*3^4*5)\) make perfect cube when multiplied by (\(2^2*3^2*5^2\))

=> i.e \((2^4*3^4*5)*(2^2*3^2*5^2)\)= \((2^2*3^2*5)^3\)

=> \(6480*\sqrt{n}\)=\((2^4*3^4*5)*\sqrt{n}\)=\((2^4*3^4*5)*(2^2*3^2*5^2)\)

=> \(\sqrt{n}\)=\((2^2*3^2*5^2)\)

=> \(\sqrt{n}\)=\((30)^2\)

=> squaring both the sides we have

=> n=\((30^2)^2\)
=> n=\(30^4\)

Therefore 'E'

Thanks
Dinesh
Re: What is the smallest positive integer n such that 6,480n^1/2   [#permalink] 25 Dec 2017, 10:32
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