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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2941
Given Kudos: 1287
GPA: 3.39
If the product of all the unique positive divisors of n, a positive in
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06 May 2021, 00:23
06 May 2021, 00:23
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If the product of all the unique positive divisors of \(n\), a positive integer which is not a perfect cube, is \(n^2\), then the product of all the unique positive divisors of \(n^2\) is
(A) \(n^3\)
(B) \(n^4\)
(C) \(n^6\)
(D) \(n^8\)
(E) \(n^9\)
(A) \(n^3\)
(B) \(n^4\)
(C) \(n^6\)
(D) \(n^8\)
(E) \(n^9\)
Re: If the product of all the unique positive divisors of n, a positive in
[#permalink]
06 May 2021, 09:10
06 May 2021, 09:10
4
Read slowly & Understand the stem.
the product of all the unique positive divisors of \(n\), a positive integer which is not a perfect cube, is \(n^2\)
Take \(n = 6\) (not a perfect cube)
divisors = 1 , 2 , 3 , 6
multiplication = 36 = \(n^2\)
Now all the divisors of 36 = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 = \(6^9\) = \(n^9\)
Answer E
the product of all the unique positive divisors of \(n\), a positive integer which is not a perfect cube, is \(n^2\)
Take \(n = 6\) (not a perfect cube)
divisors = 1 , 2 , 3 , 6
multiplication = 36 = \(n^2\)
Now all the divisors of 36 = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 = \(6^9\) = \(n^9\)
Answer E
Re: If the product of all the unique positive divisors of n, a positive in
[#permalink]
28 Jul 2026, 06:36
28 Jul 2026, 06:36
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