BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

If the greatest common divisor

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Fri Jan 12, 2018 3:42 am
lheiannie07 wrote:If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, what is the value of n?

A. 4
B. 5
C. 6
D. 7
E. 3
We can PLUG IN THE ANSWERS, which represent the value of n.
120 = 2*3*4*5 = 5!.
Since 5! is the GCF of the three given values, (n-1)! must be at least 5!.
Eliminate A, B and E, which will render too small a value for (n-1)!.

D: n=7
Here, (n+2)! = 9!, (n-1)! = 6!, and (n+4)! = 11!, with the result that all three values are divisible by 6!.
Since the GCF must be less than 6!, eliminate D.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pm
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:511 members
GMAT Score:800

by [email protected] » Fri Jan 12, 2018 11:37 am
Hi lheiannie07,

We're told that the greatest common divisor of (N+2)!, (N-1)!, and (N+4)! is 120. We're asked for the value of N. This question can be approached in a number of different ways. Here's how you can use a bit of logic and Prime Factorization to get the answer.

To start, it's worth noting that a larger factorial will be divisible by ALL of the factors that appear in a smaller factorial. For example: 4! = 24 and 3! = 6. ALL of the factors of 3! (re: 1, 2, 3 and 6) are also factors of 4! (re: 1, 2, 3, 4, 6, 8, 12, and 24). Thus, (N+2)! and (N+4)! are both bigger than (N-1)! so they will both be divisible by the same factors as (N-1)! This is meant to say that to get the answer, we just have to make sure that we make the N as small as possible AND that (N-1)! is divisible by 120.

If we 'prime factor' 120, we get....
120 =
(10)(12) =
(2)(5)(2)(2)(3)

For (N-1)! to be divisible by 120, it has to be divisible by 5... which means that '5' has to be a part of the factorial calculation. The smallest value for N in which that will happen is when N=6:
(6-1)! = 5! = (5)(4)(3)(2)(1)

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Moderator
Posts: 7187
Joined: Thu Sep 07, 2017 4:43 pm
Followed by:23 members

by BTGmoderatorDC » Sun Jan 14, 2018 4:38 pm
GMATGuruNY wrote:
lheiannie07 wrote:If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, what is the value of n?

A. 4
B. 5
C. 6
D. 7
E. 3
We can PLUG IN THE ANSWERS, which represent the value of n.
120 = 2*3*4*5 = 5!.
Since 5! is the GCF of the three given values, (n-1)! must be at least 5!.
Eliminate A, B and E, which will render too small a value for (n-1)!.

D: n=7
Here, (n+2)! = 9!, (n-1)! = 6!, and (n+4)! = 11!, with the result that all three values are divisible by 6!.
Since the GCF must be less than 6!, eliminate D.

The correct answer is C.
Thanks a lot!
Top
Moderator
Posts: 7187
Joined: Thu Sep 07, 2017 4:43 pm
Followed by:23 members

by BTGmoderatorDC » Sun Jan 14, 2018 4:43 pm
[email protected] wrote:Hi lheiannie07,

We're told that the greatest common divisor of (N+2)!, (N-1)!, and (N+4)! is 120. We're asked for the value of N. This question can be approached in a number of different ways. Here's how you can use a bit of logic and Prime Factorization to get the answer.

To start, it's worth noting that a larger factorial will be divisible by ALL of the factors that appear in a smaller factorial. For example: 4! = 24 and 3! = 6. ALL of the factors of 3! (re: 1, 2, 3 and 6) are also factors of 4! (re: 1, 2, 3, 4, 6, 8, 12, and 24). Thus, (N+2)! and (N+4)! are both bigger than (N-1)! so they will both be divisible by the same factors as (N-1)! This is meant to say that to get the answer, we just have to make sure that we make the N as small as possible AND that (N-1)! is divisible by 120.

If we 'prime factor' 120, we get....
120 =
(10)(12) =
(2)(5)(2)(2)(3)

For (N-1)! to be divisible by 120, it has to be divisible by 5... which means that '5' has to be a part of the factorial calculation. The smallest value for N in which that will happen is when N=6:
(6-1)! = 5! = (5)(4)(3)(2)(1)

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Thanks a lot!
Top
Senior | Next Rank: 100 Posts
Posts: 82
Joined: Mon Jan 15, 2018 2:01 am

TTT

by DrMaths » Mon Jan 15, 2018 4:06 am
Note that 120 = 5 x 4 x 3 x 2 x 1 = 5!
Also, (n-1)! is the smallest value that requires the greatest divisor, so we don't want anything bigger than that.
So, if we let (n-1)! = 5!, we get n = 6.
Answer C
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8086
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Mon Aug 05, 2019 6:16 pm
BTGmoderatorDC wrote:If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, what is the value of n?

A. 4
B. 5
C. 6
D. 7
E. 3

Can some experts show me solution in this problem?

OA C

We need the following fact:

If n < m, then n! is a divisor of m!.

For example, 4! is a divisor of 5! and 3! is a divisor of 8!. Now, since 5! = 120, we need n to be at least 6, in order to make (n - 1)! divisible by 5!. So this eliminates choices A, B, and E. Let's check the remaining two choices:

If n = 6, then (n+2)! = 8!, (n-1)! = 5!, and (n+4)! = 9!. Since 5! Is the smallest of them, 5! = 120 would be the greatest common divisor.

If n = 7, then (n+2)! = 9!, (n-1)! = 6!, and (n+4)! = 10!. Since 6! Is the smallest of them, 6! = 720 would be the greatest common divisor.

Therefore, we see that n must be 6.

Alternate Solution:

Since (n - 1)! is a divisor of both (n + 2)! and (n + 4)!; GCD of (n + 4)!, (n + 2)! and (n - 1)! is (n - 1)!. We are told that the GCD of these three expressions is 120, thus (n - 1)! = 120. Since 120 = 5!, we have n - 1 = 5 and so, n = 6.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Post new topic Post Reply
• Page 1 of 1