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

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

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19 Mar 2014, 07:22
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Question Stats:

64% (01:46) correct 36% (01:26) wrong based on 308 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
[Reveal] Spoiler: OA

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

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19 Mar 2014, 07:23
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 $$900^2=30^4$$.

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

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19 Mar 2014, 08:15
1
KUDOS
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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19 Mar 2014, 20:13
5
KUDOS
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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19 Mar 2014, 21:59
2
KUDOS
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 $$n^2$$ = $$3^4$$*$$5^4$$*$$2^4$$ = $$(3*5*2)^4$$ = $$30^4$$

Difficulty level - 600
Time Taken - 1:09
Math Expert
Joined: 02 Sep 2009
Posts: 44655
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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24 Mar 2014, 02:44
1
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Expert's post
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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Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]

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19 Dec 2015, 11:53
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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03 Dec 2016, 21:02
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

Great Question .

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

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25 Dec 2017, 11:32
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, 11:32
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