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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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

Show Tags

05 Feb 2014, 00:07

1

This post received 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.

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

Show Tags

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)
_________________

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.

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

Show Tags

05 May 2014, 10:37

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.

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).

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

Show Tags

05 May 2014, 12:01

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).

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

Show Tags

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.
_________________

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

Show Tags

13 Aug 2015, 16:14

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?

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).
_________________

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.

gmatclubot

Re: Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
20 Oct 2016, 06:42

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...