GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 21 Oct 2018, 18:29

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Senior Manager
Senior Manager
avatar
Joined: 10 Apr 2012
Posts: 269
Location: United States
Concentration: Technology, Other
GPA: 2.44
WE: Project Management (Telecommunications)
GMAT ToolKit User Premium Member
Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 04 Feb 2014, 17:15
4
21
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

55% (02:27) correct 45% (02:29) wrong based on 358 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.
Most Helpful Expert Reply
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8399
Location: Pune, India
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 05 Feb 2014, 09:38
6
3
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)
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

General Discussion
Magoosh GMAT Instructor
User avatar
G
Joined: 28 Dec 2011
Posts: 4493
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 04 Feb 2014, 18:33
5
4
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


Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Manager
Manager
User avatar
Joined: 18 Nov 2013
Posts: 68
Location: India
GMAT Date: 12-26-2014
WE: Information Technology (Computer Software)
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 05 Feb 2014, 01:07
1
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
Intern
avatar
Joined: 23 Jan 2014
Posts: 14
GMAT 1: 780 Q51 V45
GPA: 4
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 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)

Intern
Intern
avatar
Joined: 21 Jan 2014
Posts: 4
GMAT ToolKit User
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 05 May 2014, 11:37
1
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
User avatar
G
Joined: 28 Dec 2011
Posts: 4493
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 05 May 2014, 12:37
2
1
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


Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Intern
Intern
avatar
Joined: 21 Jan 2014
Posts: 4
GMAT ToolKit User
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 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!!!
Current Student
avatar
Joined: 09 Aug 2015
Posts: 85
GMAT 1: 770 Q51 V44
GPA: 2.3
Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 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
User avatar
P
Joined: 16 Oct 2010
Posts: 8399
Location: Pune, India
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 17 Aug 2015, 10:13
3
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

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Board of Directors
User avatar
P
Joined: 17 Jul 2014
Posts: 2657
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
GMAT ToolKit User Premium Member Reviews Badge
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 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.

answer is D.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50009
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 30 Dec 2017, 02:38
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.


VERITAS PREP OFFICIAL SOLUTION:

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)
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12694
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

Show Tags

New post 22 Mar 2018, 10:39
Hi All,

This question provides a great opportunity to TEST Values, but the real "secret" to it is in understanding how factorials and division "work." This is a relatively rare issue on the GMAT and the explanation is going to be a bit long-winded.

We're told that N > 5 and we're asked for the remainder when (N! + N + 1) is divided by (N+1).

Fact 1: (N+2) is prime.

Let's TEST N = 9

(9! +9+1) / (9+1) = (9! + 10)/10

This can be "split" into…
9!/10 + 10/10

9! = 9x8x7x6x5x4x3x2x1 = (2x5)x(9x8x7x6x4x3x1)

9!/10 has no remainder and 10/10 has no remainder, so (9! + 10)/10 has a remainder of 0

With the limitations provided by Fact 1, the remainder will ALWAYS be 0. Here's why:

Since N + 2 = prime and N > 5, then…
N MUST be odd….
N+1 MUST be even…..
N! MUST be even….(because there's a "2" in the sequence)
N! MUST be a multiple of (N+1)….(because there's a "2" and the "odd number" that you need for 2(odd) = (N+1))

(N! + N+1)/(N+1) can always be split into this….

N!/(N+1) + (N+1)/(N+1)

So… N!/(N+1) has no remainder and (N+1)/(N+1) = 1 and has no remainder.
Fact 1 ALWAYS provides a remainder of 0
Fact 1 is SUFFICIENT.

Fact 2: (N-2) is prime

Since N > 5 and Fact 2 tells us that N = odd, this creates the same circumstances as in Fact 1.
Fact 2 is SUFFICIENT.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

GMAT Club Bot
Re: Given n>5 , when (n!+n+1) is divided by (n+1) &nbs [#permalink] 22 Mar 2018, 10:39
Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.