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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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19 Mar 2014, 06:22
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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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Re: What is the smallest positive integer n such that 6,480n^1/2  [#permalink]

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19 Mar 2014, 19:13
8
3
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, 06:23
1
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, 07:15
1
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, 20:59
2
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
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Re: What is the smallest positive integer n such that 6,480n^1/2  [#permalink]

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24 Mar 2014, 01:44
1
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, 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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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

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

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07 May 2018, 10:34
Bunuel wrote:
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 $$900^2=30^4$$.

Hi Bunuel, i cant underastand the wording of this sentence must complete the powers of 2, 3 and 5 to a multiple of 3, .... to a multiple of 3 ...how can i interpret this info ? please do advise
thank you !

by the way is there a rule .... for example for a number $$xxxxxxx$$ to be divisible by number $$yyyyyyy$$ $$y$$ should have same number of prime numbers ? can you please refresh my memory with your sharp words like a clear sky lighting

pushpitkc hello may be you can explain the above
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What is the smallest positive integer n such that 6,480n^1/2  [#permalink]

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08 May 2018, 01:25
1
dave13 wrote:
Bunuel wrote:
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 $$900^2=30^4$$.

Hi Bunuel, i cant underastand the wording of this sentence must complete the powers of 2, 3 and 5 to a multiple of 3, .... to a multiple of 3 ...how can i interpret this info ? please do advise
thank you !

by the way is there a rule .... for example for a number $$xxxxxxx$$ to be divisible by number $$yyyyyyy$$ $$y$$ should have same number of prime numbers ? can you please refresh my memory with your sharp words like a clear sky lighting

pushpitkc hello may be you can explain the above

Hey dave13

In order for a number to be a perfect cube, you need to have sets of 3 prime numbers.

6480, when prime-factorized is $$2^4*3^4*5$$. In order for this number to be a perfect cube,
we need to multiply it with $$2^2*3^2*5^2 = 4*9*25 = 900$$ because then we will have 2
sets of $$2^3, 3^3,$$ and 1 set of $$5^3$$ and you will have a perfect cube.

$$900 = 30^2$$. Since, we need to multiply $$\sqrt{n}$$ with 6480 to make it a perfect cube, the value of n is $$(30^2)^2 = 30^4$$

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

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09 May 2018, 15:51
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

First, let’s factor 6,480 into primes; we have:

6480 = 10 x 648 = 10 x 9 x 72 = 10 x 9 x 9 x 8 = 2 x 5 x 3^2 x 3^2 x 2^3 = 2^4 x 3^4 x 5^1

Recall that a perfect cube requires that the exponents of all the prime factors be multiples of 3. Note that we currently have 6480 = 2^4 x 3^4 x 5^1, which means that the square root of n must be 2^2 x 3^2 x 5^2. Therefore, we see that the square root of n = 30^2, so n must be (30^2)^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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10 May 2018, 10:49
pushpitkc wrote:
dave13 wrote:
Bunuel wrote:
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 $$900^2=30^4$$.

Hi Bunuel, i cant underastand the wording of this sentence must complete the powers of 2, 3 and 5 to a multiple of 3, .... to a multiple of 3 ...how can i interpret this info ? please do advise
thank you !

by the way is there a rule .... for example for a number $$xxxxxxx$$ to be divisible by number $$yyyyyyy$$ $$y$$ should have same number of prime numbers ? can you please refresh my memory with your sharp words like a clear sky lighting

pushpitkc hello may be you can explain the above

Hey dave13

In order for a number to be a perfect cube, you need to have sets of 3 prime numbers.

6480, when prime-factorized is $$2^4*3^4*5$$. In order for this number to be a perfect cube,
we need to multiply it with $$2^2*3^2*5^2 = 4*9*25 = 900$$ because then we will have 2
sets of $$2^3, 3^3,$$ and 1 set of $$5^3$$ and you will have a perfect cube.

$$900 = 30^2$$. Since, we need to multiply $$\sqrt{n}$$ with 6480 to make it a perfect cube, the value of n is $$(30^2)^2 = 30^4$$

Hope this helps you!

hey pushpitkc

you say

"In order for a number to be a perfect cube, you need to have sets of 3 prime numbers." For example is 27 a perfect cube = 27 has thees 3s $$3^3$$ am i correct ?

or what does "sets of 3 prime numbers" mean ?

How about perfect square ? for instance 25 has 2 prime numbers - does it meant the same as two sets of prime numbers ?
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Re: What is the smallest positive integer n such that 6,480n^1/2  [#permalink]

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10 May 2018, 10:54
1
dave13 wrote:

hey pushpitkc

you say

"In order for a number to be a perfect cube, you need to have sets of 3 prime numbers." For example is 27 a perfect cube = 27 has thees 3s $$3^3$$ am i correct ?

or what does "sets of 3 prime numbers" mean ?

How about perfect square ? for instance 25 has 2 prime numbers - does it meant the same as two sets of prime numbers ?

Hi dave13

That was what I meant.
To illustrate with an example, 216 (when prime factorized) gives us 2^3 and 3^3
So, we have two sets of 3 prime numbers, making 216 a perfect cube.

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

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10 May 2018, 13:06
pushpitkc wrote:
dave13 wrote:
Bunuel wrote:
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 $$900^2=30^4$$.

Hi Bunuel, i cant underastand the wording of this sentence must complete the powers of 2, 3 and 5 to a multiple of 3, .... to a multiple of 3 ...how can i interpret this info ? please do advise
thank you !

by the way is there a rule .... for example for a number $$xxxxxxx$$ to be divisible by number $$yyyyyyy$$ $$y$$ should have same number of prime numbers ? can you please refresh my memory with your sharp words like a clear sky lighting

pushpitkc hello may be you can explain the above

Hey dave13

In order for a number to be a perfect cube, you need to have sets of 3 prime numbers.

6480, when prime-factorized is $$2^4*3^4*5$$. In order for this number to be a perfect cube,
we need to multiply it with $$2^2*3^2*5^2 = 4*9*25 = 900$$ because then we will have 2
sets of $$2^3, 3^3,$$ and 1 set of $$5^3$$ and you will have a perfect cube.

$$900 = 30^2$$. Since, we need to multiply $$\sqrt{n}$$ with 6480 to make it a perfect cube, the value of n is $$(30^2)^2 = 30^4$$

Hope this helps you!

pushpitkc YAY! i have one question

you say: "when prime-factorized is $$2^4*3^4*5$$. In order for this number to be a perfect cube,
we need to multiply it with $$2^2*3^2*5^2 = 4*9*25 = 900$$"

so when i multiply $$2^4*3^4*5$$ by $$2^2*3^2*5^2$$ = i get this:

$$2^4*2^2 = 2^6$$

$$3^4*3^2 = 3^6$$

$$5^1*5^2 = 5^3$$

$$2^6*3^6*5^3$$

so how do you get this "then we will have 2
sets of $$2^3, 3^3,$$ and 1 set of $$5^3$$"

i mean you have $$2^3$$ , $$3^3$$and $$5^3$$ whereas i get $$2^6,$$ $$3^6$$, $$5^3$$

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

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10 May 2018, 23:58
1
dave13 wrote:

pushpitkc YAY! i have one question

you say: "when prime-factorized is $$2^4*3^4*5$$. In order for this number to be a perfect cube,
we need to multiply it with $$2^2*3^2*5^2 = 4*9*25 = 900$$"

so when i multiply $$2^4*3^4*5$$ by $$2^2*3^2*5^2$$ = i get this:

$$2^4*2^2 = 2^6$$

$$3^4*3^2 = 3^6$$

$$5^1*5^2 = 5^3$$

$$2^6*3^6*5^3$$

so how do you get this "then we will have 2
sets of $$2^3, 3^3,$$ and 1 set of $$5^3$$"

i mean you have $$2^3$$ , $$3^3$$and $$5^3$$ whereas i get $$2^6,$$ $$3^6$$, $$5^3$$

can you please explain

Hi dave13

This is a basic rule of exponents $$a^{m+n} = a^m * a^n$$

Coming back to the question, where you have $$2^6$$ and $$3^6$$, we can convert this
$$2^6 = 2^{3+3} = 2^3*2^3$$ | $$3^6 = 3^{3+3} = 3^3*3^3$$

This is why I wrote that we have two sets of 2^3 and 3^3. Hope this clears your confusion.
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What is the smallest positive integer n such that 6,480n^1/2   [#permalink] 10 May 2018, 23:58
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