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What is the smallest positive integer n such that 6,480n^1/2 19 Mar 2014, 07:22
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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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What is the smallest positive integer n such that 6,480n^1/2 19 Mar 2014, 07:23
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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 the 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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What is the smallest positive integer n such that 6,480n^1/2 19 Mar 2014, 08:15
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Bunuel
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
Show more

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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What is the smallest positive integer n such that 6,480n^1/2 19 Mar 2014, 20:13
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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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What is the smallest positive integer n such that 6,480n^1/2 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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What is the smallest positive integer n such that 6,480n^1/2 24 Mar 2014, 02:44
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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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What is the smallest positive integer n such that 6,480n^1/2 19 Dec 2015, 11:53
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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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What is the smallest positive integer n such that 6,480n^1/2 03 Dec 2016, 21:02
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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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What is the smallest positive integer n such that 6,480n^1/2 25 Dec 2017, 11:32
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Bunuel
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
Show more

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 07 May 2018, 11:34
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Bunuel
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.
Show more

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 08 May 2018, 02:25
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Bunuel
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 :-)
Show more

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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What is the smallest positive integer n such that 6,480n^1/2 09 May 2018, 16:51
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Bunuel
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
Show more


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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What is the smallest positive integer n such that 6,480n^1/2 10 May 2018, 11:49
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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!
Show more


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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What is the smallest positive integer n such that 6,480n^1/2 10 May 2018, 11:54
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dave13



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 ?
Show more

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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What is the smallest positive integer n such that 6,480n^1/2 10 May 2018, 14:06
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Bunuel
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!
Show more

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 11 May 2018, 00:58
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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 :)
Show more

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 22 Jun 2025, 11:42
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Is someone able to explain how this rule works?
Bunuel
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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What is the smallest positive integer n such that 6,480n^1/2 22 Jun 2025, 13:27
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noodlebob
Is someone able to explain how this rule works?
Bunuel
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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When we find a cube of a number, we raise it to the power of 3; example,

cube of 2 =\( 2^3\),
cube of 10 = \(10^3 = (5*2)^3 = 5^3*2^3\),
cube of 12 = \(12^3 = (2^2*3)^3 = 2^6*3^3\)
cube of 27 = \(27^3 = (3^3)^3 = 3^9\)

Notice all the cubes above are nothing but base prime factorizations raised to the multiple of 3
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