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M30-07

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M30-07  [#permalink]

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New post 16 Sep 2014, 01:45
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A
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E

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  65% (hard)

Question Stats:

62% (02:03) correct 38% (01:57) wrong based on 142 sessions

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Re M30-07  [#permalink]

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New post 16 Sep 2014, 01:45
Official 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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New post 05 Dec 2014, 11:30
Bunuel I need a more in-depth explanation for this. Like where do you even get 2^2*3^2*5^2 and why is this relevant? And what do you mean by "complete the powers of 2,3, and 5 to equal a multiple of 3" and why does this matter? How does this end up getting the cube root? HELP!!
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New post 06 Dec 2014, 06:18
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scottrsmiley wrote:
Bunuel I need a more in-depth explanation for this. Like where do you even get 2^2*3^2*5^2 and why is this relevant? And what do you mean by "complete the powers of 2,3, and 5 to equal a multiple of 3" and why does this matter? How does this end up getting the cube root? HELP!!


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Re: M30-07  [#permalink]

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New post 19 Apr 2016, 02:39
6,480=2^4∗3^4∗5* root n

Now as we know just like for perfect square where we expect all the prime factors powers to be even [ multiple of 2 ].
Similarly, for perfect cube all the prime factors should be a multiple of 3.
here to make the above one to a perfect cube we need to multiply, 2^2*3^2*5^2 = 900. [ because number 2's are only 4, etc..]

So we need to have root n = 900, or root n = 30^2. There for n = 30^4
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Re: M30-07  [#permalink]

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New post 20 Jul 2016, 19:16
Is there an easy way to "break" apart 6480 to quickly find the prime factorization? The way I do it seems so time consuming!
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New post 10 Aug 2016, 15:25
If you just keep dividing by the most obvious prime factors, it doesn't take long. Eg., by 10 (5,2), then by 2 repeatedly until 81, which is 3^4. Took me about 20 seconds
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M30-07  [#permalink]

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New post 29 Jun 2018, 00:00
dmaze01 wrote:
Is there an easy way to "break" apart 6480 to quickly find the prime factorization? The way I do it seems so time consuming!


dmaze01

I am not certain if there is a perfect quick way to break it down. I split the number: 648 * 10.
This way I can note down 5*2 for 10 and start the prime factorization for 648 by picking the biggest prime factor that divides 648.

Hope this helps.
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M30-07  [#permalink]

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New post 04 Jan 2019, 09:22
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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. \(30^4\)



For what its worth, I had a very similar problem on the GMAT when I took it.

Here's my solution:

For something to be a perfect cube, all prime factors must have a power that is a multiple of 3.

Start by factoring 6,480
\(=80*81\)
\(=2^3*10*3^4\)
\(=2^4*3^4*5\)

Okay, so the smallest solution will move powers of 4 to 6 and the power of 1 to 3
I.E. we need \(2^2*3^2*5^2\)
= \(900\)

Thus,
\(\sqrt{n} = 900\)
\(\sqrt{n} = 30^2\)
\(n = 30^4\)

ANSWER E
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M30-07   [#permalink] 04 Jan 2019, 09:22
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