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  Choose one
A) n^3 B) n^4 C) n^6 D) n^8 E) n^9
the product of factors of a number 'n' would be (n)^(no.of factors/2).given the product of all the unique divisors is n^2.so we can conclude no.of factors for the given number is 4.such numbers would be of the form (p1)^1*(p2)^1 where p1 and p2 are prime numbers.   n=(p1)^1*(p2)^1   n^2=(p1)^2*(p2)^2 .   product of all the factors of n^2 would be (n^2)^(no.of factors of n^2/2)=(n^2)^(3*3/2)=n^9   ANSWn^9R:n^9
<result>E</result>
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