January 17, 2019 January 17, 2019 08:00 AM PST 09:00 AM PST Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL. January 19, 2019 January 19, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 10 Apr 2012
Posts: 268
Location: United States
Concentration: Technology, Other
GPA: 2.44
WE: Project Management (Telecommunications)

Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
Show Tags
04 Feb 2014, 16:15
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.
Official Answer and Stats are available only to registered users. Register/ Login.




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India

Re: Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
Show Tags
05 Feb 2014, 08:38
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. 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 (n2) is prime, (n1) 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 >




Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4486

Re: Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
Show Tags
04 Feb 2014, 17:33
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
Joined: 18 Nov 2013
Posts: 63
Location: India
GMAT Date: 12262014
WE: Information Technology (Computer Software)

Re: Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
Show Tags
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: 14
GPA: 4

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



Intern
Joined: 21 Jan 2014
Posts: 4

Re: Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
Show Tags
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 (n2) is prime, (n1) 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: 4486

Re: Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
Show Tags
05 May 2014, 11:37
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
Joined: 21 Jan 2014
Posts: 4

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). Does all this make sense? Mike Now I get it Thank you for the explanation!!!



Manager
Joined: 09 Aug 2015
Posts: 84
GPA: 2.3

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 n2 is the prime 2?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India

Re: Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
Show Tags
17 Aug 2015, 09:13
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 n2 is the prime 2? Yes, it is to exclude the case where (n2) 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 >



Board of Directors
Joined: 17 Jul 2014
Posts: 2600
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
Show Tags
20 Oct 2016, 06:42
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. n2 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
Joined: 02 Sep 2009
Posts: 52230

Re: Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
Show Tags
30 Dec 2017, 01:38
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 (n2) is prime, (n1) 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? Extrahard Quant Tests with Brilliant Analytics



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13331
Location: United States (CA)

Re: Given n>5 , when (n!+n+1) is divided by (n+1)
[#permalink]
Show Tags
22 Mar 2018, 09:39
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 longwinded. 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: (N2) 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
CoFounder & 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!*****




Re: Given n>5 , when (n!+n+1) is divided by (n+1) &nbs
[#permalink]
22 Mar 2018, 09:39






