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E
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Difficulty:
95%
(hard)
Question Stats:
25% (02:51) correct
75%
(02:15)
wrong
based on 3900
sessions
History
Date
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Not Attempted Yet
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) \(n^9\)
02 Feb 2011, 20:57
Kudos
Bookmarks
I agree with E: n^9.
Another way to look at it (using the same methodology outlined by Bunuel in his post):
We know that the product of all the divisors of n is equivalent to n^2. Therefore, n = x * y, where x and y are both prime positive and distinct integers (as outlined by the conditions set in the original question, and explained in the previous post).
We need to determine, the product of all the divisors of n^2, which we know to be equivalent to:
\(n^2 = n * n\)
\(n^2 = (x * y)(x * y)\)
Since both x and y are known to be prime numbers, the divisors of n^2 are therefore:
\(1, x, y, xy, x^2, y^2, x^2y, xy^2,\) and \(x^2y^2\)
The product of these divisors is therefore:
\((1)(x)(y)(xy)(x^2)(y^2)(x^2y)(xy^2)(x^2y^2)\)
\(= x^9y^9\)
\(= n^9\)
Another way to look at it (using the same methodology outlined by Bunuel in his post):
We know that the product of all the divisors of n is equivalent to n^2. Therefore, n = x * y, where x and y are both prime positive and distinct integers (as outlined by the conditions set in the original question, and explained in the previous post).
We need to determine, the product of all the divisors of n^2, which we know to be equivalent to:
\(n^2 = n * n\)
\(n^2 = (x * y)(x * y)\)
Since both x and y are known to be prime numbers, the divisors of n^2 are therefore:
\(1, x, y, xy, x^2, y^2, x^2y, xy^2,\) and \(x^2y^2\)
The product of these divisors is therefore:
\((1)(x)(y)(xy)(x^2)(y^2)(x^2y)(xy^2)(x^2y^2)\)
\(= x^9y^9\)
\(= n^9\)
02 Feb 2011, 12:46
Kudos
Bookmarks
gmatpapa
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.
