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What is the smallest positive integer n such that 6,480n^1/2
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19 Mar 2014, 07:22
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Re: What is the smallest positive integer n such that 6,480n^1/2
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19 Mar 2014, 20:13
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
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19 Mar 2014, 07:23



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Re: What is the smallest positive integer n such that 6,480n^1/2
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19 Mar 2014, 08:15
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
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19 Mar 2014, 21:59
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
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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
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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



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What is the smallest positive integer n such that 6,480n^1/2
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07 May 2018, 11:34
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 \(900^2=30^4\).
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



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What is the smallest positive integer n such that 6,480n^1/2
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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 dave13In order for a number to be a perfect cube, you need to have sets of 3 prime numbers. 6480, when primefactorized 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
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09 May 2018, 16:51
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. Answer: E
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Re: What is the smallest positive integer n such that 6,480n^1/2
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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 dave13In order for a number to be a perfect cube, you need to have sets of 3 prime numbers. 6480, when primefactorized 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
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10 May 2018, 11:54
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 dave13That 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
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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 dave13In order for a number to be a perfect cube, you need to have sets of 3 prime numbers. 6480, when primefactorized 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 primefactorized 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
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11 May 2018, 00:58
dave13 wrote: pushpitkc YAY! i have one question you say: "when primefactorized 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 dave13This 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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