Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 26 May 2017, 15:57

### 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Given n>5 , when (n!+n+1) is divided by (n+1)

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 10 Apr 2012
Posts: 276
Location: United States
Concentration: Technology, Other
GPA: 2.44
WE: Project Management (Telecommunications)
Followers: 5

Kudos [?]: 907 [1] , given: 325

Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]

### Show Tags

04 Feb 2014, 17:15
1
KUDOS
12
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

51% (02:51) correct 49% (01:55) wrong based on 230 sessions

### HideShow timer Statistics

Given n>5 , when (n!+n+1) is divided by (n+1) , what is the remainder ?

(1) (n+2) is a prime number.

(2) (n−2) is a prime number.
[Reveal] Spoiler: OA
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4041
Followers: 1417

Kudos [?]: 6783 [3] , given: 84

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]

### Show Tags

04 Feb 2014, 18:33
3
KUDOS
Expert's post
3
This post was
BOOKMARKED
guerrero25 wrote:
Given n>5 , when (n!+n+1) is divided by (n+1) , what is the remainder ?

(1) (n+2) is a prime number.

(2) (n−2) is a prime number.

This is a truly brilliant question, and I am very happy to help.

When when (n!+n+1) is divided by (n+1), (n + 1) obviously divides evenly into the (n + 1), so the question is: what is the remainder when we divide (n!) by (n+1)?

Well, if n > 5, then the only odd numbers available are prime. Whether (n + 2) or (n - 2) is prime, that means that n must be odd. If n is odd, (n + 1) is even, and therefore is product of 2 and something smaller than n. That means, both factors of (n + 1) must be contained in (n!), so (n + 1) divides evenly into (n!) with a remainder of zero. Each statement is sufficient on its own. OA = (D).

For example, let n = 21. Both n - 2 = 19 and n + 2 = 23 are prime. We know that (n + 1) = 22 is even: thus, 22 = 2*11, and of course, (21!) must contain every factor from 1 to 21, so it will include both a factor of 2 and factor of 11. Therefore, (21!) is necessarily a multiple of 22, and were we to divide, we would get a remainder of zero.

Does all this make sense?
Mike
_________________

Mike McGarry
Magoosh Test Prep

Manager
Joined: 18 Nov 2013
Posts: 81
Location: India
GMAT Date: 12-26-2014
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 19 [1] , given: 7

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]

### Show Tags

05 Feb 2014, 01:07
1
KUDOS
mikemcgarry wrote:
guerrero25 wrote:
Given n>5 , when (n!+n+1) is divided by (n+1) , what is the remainder ?

(1) (n+2) is a prime number.

(2) (n−2) is a prime number.

This is a truly brilliant question, and I am very happy to help.

When when (n!+n+1) is divided by (n+1), (n + 1) obviously divides evenly into the (n + 1), so the question is: what is the remainder when we divide (n!) by (n+1)?

Well, if n > 5, then the only odd numbers available are prime. Whether (n + 2) or (n - 2) is prime, that means that n must be odd. If n is odd, (n + 1) is even, and therefore is product of 2 and something smaller than n. That means, both factors of (n + 1) must be contained in (n!), so (n + 1) divides evenly into (n!) with a remainder of zero. Each statement is sufficient on its own. OA = (D).

For example, let n = 21. Both n - 2 = 19 and n + 2 = 23 are prime. We know that (n + 1) = 22 is even: thus, 22 = 2*11, and of course, (21!) must contain every factor from 1 to 21, so it will include both a factor of 2 and factor of 11. Therefore, (21!) is necessarily a multiple of 22, and were we to divide, we would get a remainder of zero.

Does all this make sense?
Mike

It sure does makes sense. I just supplemented values for n in the equation to get to the answer. I hope that's okay while solving such questions.
Intern
Joined: 23 Jan 2014
Posts: 16
GMAT 1: 780 Q51 V45
GPA: 4
Followers: 0

Kudos [?]: 12 [0], given: 0

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]

### Show Tags

05 Feb 2014, 08:45
Quote:

It sure does makes sense. I just supplemented values for n in the equation to get to the answer. I hope that's okay while solving such questions.

doing this way will not always work and might cause you to make false assessment (if something is true for a given value of n does not mean it is true for all n)
_________________

GMAT score: 780 (51 quant, 45 verbal, 5.5 AWA, 8 IR)

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7376
Location: Pune, India
Followers: 2288

Kudos [?]: 15127 [5] , given: 224

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]

### Show Tags

05 Feb 2014, 09:38
5
KUDOS
Expert's post
2
This post was
BOOKMARKED
guerrero25 wrote:
Given n>5 , when (n!+n+1) is divided by (n+1), what is the remainder?

(1) (n+2) is a prime number.

(2) (n−2) is a prime number.

I would like to weigh in here since it is one of my little creations!
The concept I wanted to test here was that if (n+1) is prime, it will not be a factor of n! because (n+1) cannot be broken down into smaller factors. On the other hand, a composite number can be broken down into 2 smaller factors and both will be included in n!. Here (n+1) is always even so it can definitely be broken down into two factors: 2 and n/2. In any case, n! will have both 2 and n/2 as factors (simultaneously).

Here is the detailed explanation:

Given (n! + n + 1)

$$\frac{n! + (n+1)}{n+1} = \frac{n!}{n+1} + 1$$

Now if n! is divisible by (n+1), then the remainder is 0. If not, then we don’t know the remainder.
If (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1). If (n+1) is prime, n! does not contain (n+1) and hence is not divisible by (n+1).

Statement I: If (n+2) is prime, (n+1) must be even and hence composite (recall that there is only one prime even number i.e. 2). Remainder must be 0. Sufficient.

Statement II: If (n-2) is prime, (n-1) and (n+1) must be even and hence composite. Remainder must be 0. Sufficient.

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 21 Jan 2014 Posts: 5 Followers: 0 Kudos [?]: 1 [1] , given: 2 Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink] ### Show Tags 05 May 2014, 11:37 1 This post received KUDOS VeritasPrepKarishma wrote: guerrero25 wrote: Given n>5 , when (n!+n+1) is divided by (n+1), what is the remainder? (1) (n+2) is a prime number. (2) (n−2) is a prime number. I would like to weigh in here since it is one of my little creations! The concept I wanted to test here was that if (n+1) is prime, it will not be a factor of n! because (n+1) cannot be broken down into smaller factors. On the other hand, a composite number can be broken down into 2 smaller factors and both will be included in n!. Here (n+1) is always even so it can definitely be broken down into two factors: 2 and n/2. In any case, n! will have both 2 and n/2 as factors (simultaneously). Here is the detailed explanation: Given (n! + n + 1) $$\frac{n! + (n+1)}{n+1} = \frac{n!}{n+1} + 1$$ Now if n! is divisible by (n+1), then the remainder is 0. If not, then we don’t know the remainder. If (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1). If (n+1) is prime, n! does not contain (n+1) and hence is not divisible by (n+1). Statement I: If (n+2) is prime, (n+1) must be even and hence composite (recall that there is only one prime even number i.e. 2). Remainder must be 0. Sufficient. Statement II: If (n-2) is prime, (n-1) and (n+1) must be even and hence composite. Remainder must be 0. Sufficient. Answer (D) Hi Karishma, could you please elaborate this part of your reasoning: If (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1) I don't really see why if n+1 is even then n! will be divisible by n+1. Thanks in advance Magoosh GMAT Instructor Joined: 28 Dec 2011 Posts: 4041 Followers: 1417 Kudos [?]: 6783 [2] , given: 84 Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink] ### Show Tags 05 May 2014, 12:37 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED NAL9 wrote: Hi Karishma, could you please elaborate this part of your reasoning: If (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1) I don't really see why if n+1 is even then n! will be divisible by n+1. Thanks in advance Dear NAL9, I saw this, so I'll respond. I have a great deal of respect for Karishma, but I can answer this question. Think about it this way. The number (73!) is a very big number, more than 100 digits. It is the product of every integer from 1 to 73, so it is automatically divisible by any integer from 1 to 73. Now, think about the next integer, 74. We want to know whether (73!), that 100+ digit number, will be divisible by 74. Well, notice that 74 = 2*37. Clearly (73!) is divisible by both 2 and by 37: therefore, it must be divisible by 2*37 = 74. In much the same way, as long as we know that n is odd and that (n + 1) is even, we know that (n + 1) can be written as a product of 2 times some other number. That other integer, (n + 1)/2, will be much smaller than n, and therefore automatically will be one of the factors included in (n!). Of course, 2 is another factor also included in (n!). Therefore, (n!) would have to be divisible by the product of these two factors, and that product is (n + 1). Does all this make sense? Mike _________________ Mike McGarry Magoosh Test Prep Intern Joined: 21 Jan 2014 Posts: 5 Followers: 0 Kudos [?]: 1 [0], given: 2 Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink] ### Show Tags 05 May 2014, 13:01 mikemcgarry wrote: NAL9 wrote: Hi Karishma, could you please elaborate this part of your reasoning: If (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1) I don't really see why if n+1 is even then n! will be divisible by n+1. Thanks in advance Dear NAL9, I saw this, so I'll respond. I have a great deal of respect for Karishma, but I can answer this question. Think about it this way. The number (73!) is a very big number, more than 100 digits. It is the product of every integer from 1 to 73, so it is automatically divisible by any integer from 1 to 73. Now, think about the next integer, 74. We want to know whether (73!), that 100+ digit number, will be divisible by 74. Well, notice that 74 = 2*37. Clearly (73!) is divisible by both 2 and by 37: therefore, it must be divisible by 2*37 = 74. In much the same way, as long as we know that n is odd and that (n + 1) is even, we know that (n + 1) can be written as a product of 2 times some other number. That other integer, (n + 1)/2, will be much smaller than n, and therefore automatically will be one of the factors included in (n!). Of course, 2 is another factor also included in (n!). Therefore, (n!) would have to be divisible by the product of these two factors, and that product is (n + 1). Does all this make sense? Mike Now I get it Thank you for the explanation!!! GMAT Club Legend Joined: 09 Sep 2013 Posts: 15466 Followers: 649 Kudos [?]: 209 [0], given: 0 Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink] ### Show Tags 14 May 2015, 04:08 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Current Student Joined: 09 Aug 2015 Posts: 95 GMAT 1: 770 Q51 V44 GPA: 2.3 Followers: 0 Kudos [?]: 22 [0], given: 6 Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink] ### Show Tags 13 Aug 2015, 17:14 This is a great question; it tests thoroughly tests your brain. is there any significance of the n>5? Is it to exclude the case where n-2 is the prime 2? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7376 Location: Pune, India Followers: 2288 Kudos [?]: 15127 [2] , given: 224 Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink] ### Show Tags 17 Aug 2015, 10:13 2 This post received KUDOS Expert's post ggurface wrote: This is a great question; it tests thoroughly tests your brain. is there any significance of the n>5? Is it to exclude the case where n-2 is the prime 2? Yes, it is to exclude the case where (n-2) could be an even prime. If we are talking about primes greater than 2, they will definitely be odd so with either statement, (n+1) will definitely be even and greater than 2 (hence, non prime). _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

CEO
Joined: 17 Jul 2014
Posts: 2510
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '19 (D)
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Followers: 26

Kudos [?]: 342 [0], given: 169

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]

### Show Tags

20 Oct 2016, 07:42
guerrero25 wrote:
Given n>5 , when (n!+n+1) is divided by (n+1) , what is the remainder ?

(1) (n+2) is a prime number.

(2) (n−2) is a prime number.

i got to D.

n>5 -> n!+n+1/(n+1) -> rewrite it: n!/(n+1) + (n+1)/(+1)
n!/(n+1) +1

we need to find n!/(n+1)

1. n+2 is prime -> n=9, n=11, n=15, n=17, etc.
let's take few examples and identify the pattern
9!
9! contains a 2 and a 5, so definitely, it will be a multiple of 10. remainder will be 0.
11!/12 -> 11!
11! contain a 3 and a 4, so when divided by 12, the remainder will be 0...same thing continues for all the options...

2. n-2 prime
n=7, 11, 17, etc.
7!/8 - remainder is 0
11!/12 - remainder is 0.

same pattern
statement 2 is sufficient.

Re: Given n>5 , when (n!+n+1) is divided by (n+1)   [#permalink] 20 Oct 2016, 07:42
Similar topics Replies Last post
Similar
Topics:
2 If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24? 3 06 May 2017, 03:02
2 For a positive integer n, what is the remainder when n(n+1) is divided 4 06 Oct 2016, 05:42
9 Is n(n+1)(n+2) divisible by 24? 3 01 Oct 2016, 23:47
4 Is n(n + 1) a multiple of 3? 4 13 Jan 2015, 14:20
5 For all integers n, n* = n(n – 1). what is the value of x* 6 20 Oct 2016, 07:24
Display posts from previous: Sort by