What is the smallest positive integer n such that 6,480n^1/2 : Problem Solving (PS)

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

19 Mar 2014, 08:15

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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]

19 Mar 2014, 21:59

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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
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]

24 Mar 2014, 02:44

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Expert Reply

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]

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]

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 .

B

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

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

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

07 May 2018, 11:34

1

Kudos

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 $900^2=30^4$.

Answer: E.

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

L

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

08 May 2018, 02:25

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 $900^2=30^4$.

Answer: E.

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!

L

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

09 May 2018, 16:51

Expert Reply

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.

Answer: E

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

10 May 2018, 11:49

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 $900^2=30^4$.

Answer: E.

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 ?

L

pushpitkc

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

10 May 2018, 11:54

1

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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!

L

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

10 May 2018, 14:06

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 $900^2=30^4$.

Answer: E.

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

L

pushpitkc

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

11 May 2018, 00:58

1

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

12 Sep 2023, 04:22

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

12 Sep 2023, 04:22

GMAT Problem Solving (PS) Questions

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