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19 Mar 2014, 07:22
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E
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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
A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4
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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
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
19 Mar 2014, 07:23
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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:
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 (which are 4, 4, and 1, respectively) to a multiple of 3. Thus the least value of \(\sqrt{n}\) must equal to \(2^2*3^2*5^2\).
Notice that in this case WILL be a perfect cube:
The least value of \(n\), therefore, is:
Answer: E.
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 (which are 4, 4, and 1, respectively) to a multiple of 3. Thus the least value of \(\sqrt{n}\) must equal to \(2^2*3^2*5^2\).
Notice that in this case WILL be a perfect cube:
- \(2^4*3^4*5*\sqrt{n}=2^4*3^4*5*(2^2*3^2*5^2)=2^6*3^6*5^3=(2^2*3^2*5)^3\).
The least value of \(n\), therefore, is:
- \((\sqrt{n})^2=n=(2^2*3^2*5^2)^2=30^4\).
Answer: E.
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