Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
25% (02:51) correct
75%
(02:15)
wrong
based on 3900
sessions
History
Date
Time
Result
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.
