If the product of all the unique positive divisors of n, a positive in

All positive integers \(n\) which equal to \(n=p_1*p_2\), where \(p_1\) and \(p_2\) are distinct primes satisfy the condition in the stem. Because the factors of \(n\) in this case would be: 1, \(p_1\), \(p_2\), and \(n\) itself, so the product of the factors will be \(1*(p_1*p_2)*n=n^2\).

(Note that if \(n=p^3\) where \(p\) is a prime number also satisfies this condition as the factors of \(n\) in this case would be 1, \(p\), \(p^2\) and \(n\) itself, so the product of the factors will be \(1*(p*p^2)*n=p^3*n=n^2\), but we are told that \(n\) is not a perfect cube, so this case is out, as well as the case \(n=1\).)

For example if \(n=6=2*3\) --> the product of all the unique positive divisors of 6 will be: \(1*2*3*6=6^2\);
Or if \(n=10=2*5\) --> the product of all the unique positive divisors of 10 will be: \(1*2*5*10=10^2\);

Now, take \(n=10\) --> \(n^2=100\) --> the product of all the unique positive divisors of \(100\) is: \(1*2*4*5*10*20*25*50*100=(2*50)*(4*25)*(5*20)*10*100=10^2*10^2*10^2*10*10^2=10^9\)

(You can do this with formula to get \((p_1)^9*(p_2)^9=n^9\) but think this way is quicker).

Answer: E.