sudhirgupta93 wrote:
Bunuel wrote:
gmatpapa wrote:
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) \(n^9\)
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.
Can someone please explain why we nave taken +ve integer n of the form n= p1 x p2?
I'm unable to comprehend as to which constraint given in the question bounds us to imply that n is a product of only 2 prime numbers (without any power)..
"the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n^2"
Had n been a prime number, it would have only 2 factors: 1 and n
Product of all factors = 1*n = n
If product of all factors is instead n^2, it means that other than 1 and n, it has 2 more factors which also multiply to give n.
All factors: 1, p1, p2, n (where p1 and p2 are two prime numbers)
Product of all factors = 1*p1*p2*n = n^2
What if p1 and p2 were not prime? e.g. 1, 4, 6, 24
Here, there are other factors too: 2, 3, 8 etc. So the product of all factors would be much more than n^2.
If there have to be only 4 factors, they must be 1, p1, p2 and n where p1 and p2 must be prime factors.
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Karishma
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