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
26 May 2017, 04:48
Kudos
Bookmarks
If the greatest common divisor of \((2n)!, (n-2)!,\) and \((n+4)!\) is 24 where n is an integer greater than 4, what is the value of n?
A. 4
B. 5
C. 6
D. 7
E. 3
A. 4
B. 5
C. 6
D. 7
E. 3
26 May 2017, 04:49
Kudos
Bookmarks
Official Solution:
If the greatest common divisor of \((2n)!, (n-2)!,\) and \((n+4)!\) is 24 where n is an integer greater than 4, what is the value of n?
A. 4
B. 5
C. 6
D. 7
E. 3
The greatest common divisor (factor) is literally the biggest factor among the common factors. Also, since n is an integer greater than 4, you get \(2n > n + 4 > n - 2\), then \((2n)! = 2n(2n-1)! = 2n(2n-1)\ldots ..(n-1)(n-2)!\), and then \((n+4)! = (n+4)(n+3)(n+2)\ldots ..(n-1)(n-2)!\).
Thus, \((n-2)!\) is the common factor of \((n+4)!\) and \((2n)!\). Eventually,
the greatest common divisor of \((2n)!, (n-2)!,\) and \((n+4)!\) is \((n-2)!\).
If so, from \((n-2)! = 24 = 4!\), you get \(n-2 = 4, n = 6.\) The answer is C.
Answer: C
If the greatest common divisor of \((2n)!, (n-2)!,\) and \((n+4)!\) is 24 where n is an integer greater than 4, what is the value of n?
A. 4
B. 5
C. 6
D. 7
E. 3
The greatest common divisor (factor) is literally the biggest factor among the common factors. Also, since n is an integer greater than 4, you get \(2n > n + 4 > n - 2\), then \((2n)! = 2n(2n-1)! = 2n(2n-1)\ldots ..(n-1)(n-2)!\), and then \((n+4)! = (n+4)(n+3)(n+2)\ldots ..(n-1)(n-2)!\).
Thus, \((n-2)!\) is the common factor of \((n+4)!\) and \((2n)!\). Eventually,
the greatest common divisor of \((2n)!, (n-2)!,\) and \((n+4)!\) is \((n-2)!\).
If so, from \((n-2)! = 24 = 4!\), you get \(n-2 = 4, n = 6.\) The answer is C.
Answer: C
Moderator:
