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# M30-07

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Math Expert
Joined: 02 Sep 2009
Posts: 55624

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16 Sep 2014, 01:45
00:00

Difficulty:

65% (hard)

Question Stats:

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

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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. $$30^4$$

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Math Expert
Joined: 02 Sep 2009
Posts: 55624

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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 $$900^2=30^4$$.

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Joined: 23 Jul 2014
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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!!
Math Expert
Joined: 02 Sep 2009
Posts: 55624

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06 Dec 2014, 06:18
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Intern
Joined: 07 Mar 2011
Posts: 23

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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
Intern
Joined: 22 Jun 2016
Posts: 8

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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!
Intern
Joined: 20 Jul 2016
Posts: 2

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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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Joined: 12 Aug 2017
Posts: 6

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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.
Intern
Joined: 05 Oct 2018
Posts: 43
Location: United States
GMAT 1: 770 Q49 V47
GPA: 3.95
WE: General Management (Other)

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04 Jan 2019, 09:22
2
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,
$$\sqrt{n} = 900$$
$$\sqrt{n} = 30^2$$
$$n = 30^4$$

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

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