Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 03 Aug 2015, 05:37

### 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:
Senior Manager
Joined: 10 Apr 2012
Posts: 281
Location: United States
Concentration: Technology, Other
GPA: 2.44
WE: Project Management (Telecommunications)
Followers: 3

Kudos [?]: 281 [0], given: 325

Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]  04 Feb 2014, 16:15
6
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

54% (02:56) correct 46% (02:03) wrong based on 114 sessions
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: 2461
Followers: 796

Kudos [?]: 3190 [3] , given: 38

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]  04 Feb 2014, 17:33
3
KUDOS
Expert's post
1
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: 84
Location: India
GMAT Date: 12-26-2014
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 15 [0], given: 7

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]  05 Feb 2014, 00:07
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: 17
GMAT 1: 780 Q51 V45
GPA: 4
Followers: 0

Kudos [?]: 9 [0], given: 0

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]  05 Feb 2014, 07:45
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: 5752
Location: Pune, India
Followers: 1445

Kudos [?]: 7614 [2] , given: 186

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]  05 Feb 2014, 08:38
2
KUDOS
Expert's post
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 [?]: 0 [0], given: 2

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]  05 May 2014, 10:37
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.

Hi Karishma,

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.

Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 2461
Followers: 796

Kudos [?]: 3190 [2] , given: 38

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]  05 May 2014, 11:37
2
KUDOS
Expert's post
NAL9 wrote:
Hi Karishma,

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.

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 [?]: 0 [0], given: 2

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]  05 May 2014, 12:01
mikemcgarry wrote:
NAL9 wrote:
Hi Karishma,

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.

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: 5723
Followers: 323

Kudos [?]: 63 [0], given: 0

Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]  14 May 2015, 03:08
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.
_________________
Re: Given n>5 , when (n!+n+1) is divided by (n+1)   [#permalink] 14 May 2015, 03:08
Similar topics Replies Last post
Similar
Topics:
4 Is n(n+1)(n+2) divisible by 24? 2 29 Apr 2015, 03:56
2 Is n(n + 1) a multiple of 3? 4 07 Sep 2013, 05:02
5 For all integers n, n* = n(n – 1). what is the value of x* 5 14 Jul 2013, 19:34
10 Given that n is an integer, is n 1 divisible by 3? 21 18 May 2010, 04:36
9 For all integers n, n* = n(n – 1). What is the value of x* w 15 20 Nov 2009, 15:24
Display posts from previous: Sort by