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

 It is currently 19 Feb 2019, 09:21

### 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 February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Prep Hour

February 20, 2019

February 20, 2019

08:00 PM EST

09:00 PM EST

Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST

February 21, 2019

February 21, 2019

10:00 PM PST

11:00 PM PST

Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.

# M27-01

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52971

### Show Tags

16 Sep 2014, 00:26
2
18
00:00

Difficulty:

95% (hard)

Question Stats:

29% (02:13) correct 71% (01:41) wrong based on 228 sessions

### HideShow timer Statistics

If $$n$$ is a positive integer and $$p$$ is a prime number, is $$p$$ a factor of $$n!$$?

(1) $$p$$ is a factor of $$(n+2)!-n!$$

(2) $$p$$ is a factor of $$\frac{(n+2)!}{n!}$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52971

### Show Tags

16 Sep 2014, 00:26
13
Official Solution:

(1) $$p$$ is a factor of $$(n+2)!-n!$$. If $$n=2$$ then $$(n+2)!-n!=22$$ and for $$p=2$$ the answer will be YES, but for $$p=11$$ the answer will be NO. Not sufficient.

(2) $$p$$ is a factor of $$\frac{(n+2)!}{n!}$$. Simplify: $$\frac{(n+2)!}{n!}=(n+1)(n+2)$$. Now, if $$n=2$$ then $$(n+1)(n+2)=12$$ and for $$p=2$$ the answer will be YES, but for $$p=3$$ the answer will be NO. Not sufficient.

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$. Now, $$(n+1)(n+2)-1$$ and $$(n+1)(n+2)$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example, 20 and 21 are consecutive integers, thus the only common factor they share is 1. So, as from (2) $$p$$ is a factor of $$(n+1)(n+2)$$ then it cannot be a factor of $$(n+1)(n+2)-1$$. Thus, in order for $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$. Sufficient.

_________________
Intern
Joined: 11 Nov 2014
Posts: 34
Concentration: Marketing, Finance
WE: Programming (Computer Software)

### Show Tags

15 Mar 2015, 16:39
Bunuel wrote:
Official Solution:

(1) $$p$$ is a factor of $$(n+2)!-n!$$. If $$n=2$$ then $$(n+2)!-n!=22$$ and for $$p=2$$ the answer will be YES, but for $$p=11$$ the answer will be NO. Not sufficient.

(2) $$p$$ is a factor of $$\frac{(n+2)!}{n!}$$. Simplify: $$\frac{(n+2)!}{n!}=(n+1)(n+2)$$. Now, if $$n=2$$ then $$(n+1)(n+2)=12$$ and for $$p=2$$ the answer will be YES, but for $$p=3$$ the answer will be NO. Not sufficient.

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$. Now, $$(n+1)(n+2)-1$$ and $$(n+1)(n+2)$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example, 20 and 21 are consecutive integers, thus the only common factor they share is 1. So, as from (2) $$p$$ is a factor of $$(n+1)(n+2)$$ then it cannot be a factor of $$(n+1)(n+2)-1$$. Thus, in order for $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$. Sufficient.

hello,
one small doubt.
After combining (1) and (2) -
will "p" always be 1 ??
Math Expert
Joined: 02 Aug 2009
Posts: 7334

### Show Tags

15 Mar 2015, 18:07
1
minhaz3333 wrote:
Bunuel wrote:
Official Solution:

(1) $$p$$ is a factor of $$(n+2)!-n!$$. If $$n=2$$ then $$(n+2)!-n!=22$$ and for $$p=2$$ the answer will be YES, but for $$p=11$$ the answer will be NO. Not sufficient.

(2) $$p$$ is a factor of $$\frac{(n+2)!}{n!}$$. Simplify: $$\frac{(n+2)!}{n!}=(n+1)(n+2)$$. Now, if $$n=2$$ then $$(n+1)(n+2)=12$$ and for $$p=2$$ the answer will be YES, but for $$p=3$$ the answer will be NO. Not sufficient.

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$. Now, $$(n+1)(n+2)-1$$ and $$(n+1)(n+2)$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example, 20 and 21 are consecutive integers, thus the only common factor they share is 1. So, as from (2) $$p$$ is a factor of $$(n+1)(n+2)$$ then it cannot be a factor of $$(n+1)(n+2)-1$$. Thus, in order for $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$. Sufficient.

hello,
one small doubt.
After combining (1) and (2) -
will "p" always be 1 ??

hi,
p will always not be 1and is not one in this case too....
combined , it tells us that the first has been broken down to n!(n+1)(n+2) and second (n+1)(n+2)-1...
now (n+1)(n+2) and (n+1)(n+2)-1 will have only 1 as factor as these two are consecutive...
p is prime and cannot be 1..
so p can be a factor of (n+1)(n+2)-1 in (2) but will be a factor of n! in (1)
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html

GMAT Expert

Intern
Joined: 01 Aug 2013
Posts: 4

### Show Tags

22 Jul 2016, 20:36
Hi Bunuel ,

In the steps below,
(1)+(2) (n+2)!−n!=n!((n+1)(n+2)−1). Now, (n+1)(n+2)−1 and (n+1)(n+2) are consecutive integers.
Could you please let me know, why did you discard the n! here, and have taken (n+1)(n+2)−1 , instead of n!((n+1)(n+2)−1). Where am I missing?

Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 52971

### Show Tags

22 Jul 2016, 23:15
1
sathishm07 wrote:
Hi Bunuel ,

In the steps below,
(1)+(2) (n+2)!−n!=n!((n+1)(n+2)−1). Now, (n+1)(n+2)−1 and (n+1)(n+2) are consecutive integers.
Could you please let me know, why did you discard the n! here, and have taken (n+1)(n+2)−1 , instead of n!((n+1)(n+2)−1). Where am I missing?

Thanks!

Please re-read the solution. I just analyzed tow multiples of n!((n+1)(n+2)−1) separately: (n+1)(n+2)−1 first and the n!.
_________________
Intern
Joined: 29 Oct 2014
Posts: 38

### Show Tags

23 Jul 2016, 01:45
Bunuel wrote:
Official Solution:

(1) $$p$$ is a factor of $$(n+2)!-n!$$. If $$n=2$$ then $$(n+2)!-n!=22$$ and for $$p=2$$ the answer will be YES, but for $$p=11$$ the answer will be NO. Not sufficient.

(2) $$p$$ is a factor of $$\frac{(n+2)!}{n!}$$. Simplify: $$\frac{(n+2)!}{n!}=(n+1)(n+2)$$. Now, if $$n=2$$ then $$(n+1)(n+2)=12$$ and for $$p=2$$ the answer will be YES, but for $$p=3$$ the answer will be NO. Not sufficient.

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$. Now, $$(n+1)(n+2)-1$$ and $$(n+1)(n+2)$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example, 20 and 21 are consecutive integers, thus the only common factor they share is 1. So, as from (2) $$p$$ is a factor of $$(n+1)(n+2)$$ then it cannot be a factor of $$(n+1)(n+2)-1$$. Thus, in order for $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$. Sufficient.

great explanation Bunuel
question too:)
Current Student
Joined: 08 Jan 2015
Posts: 77

### Show Tags

26 Aug 2016, 23:56
I used the same logic for 1) and 2) combined, but a bit different for 1) and 2) separated:

1) p is a factor of (n+2)! - n! and this means that p is a factor of the following expression:
(n+2)! - n! = n!*(n+1)*(n+2) - n! = n! * [(n+1)*(n+2) - 1]
Then it means that p is a factor of either n! or [(n+1)*(n+2) - 1], but not both of them. Actually, I don't know how to prove that it can't be a factor of both, but the expression is pretty weird and my gut feeling is that the restriction is that these number are somehow co-primes... Any thoughts?

2) p is a factor of (n+2)!/n! and this means that p is a factor of the following expression:
(n+2)!/n! = [n!*(n+1)*(n+2)]/n! = (n+1)*(n+2)
This means that p is a factor of either (n+1) or (n+2), which are co-prime, thus it will give two different answers for the question.

3) Combined, as Bunuel mentions, the only way p to be a factor of both expression is to be the factor of n!, otherwise it should be the factor of co-primes and this is not feasible.
Current Student
Joined: 31 Jan 2016
Posts: 24
Concentration: Finance, Statistics
GMAT 1: 690 Q47 V37
GPA: 3.6

### Show Tags

06 Sep 2016, 09:09
Bunuel wrote:
Official Solution:

(1) $$p$$ is a factor of $$(n+2)!-n!$$. If $$n=2$$ then $$(n+2)!-n!=22$$ and for $$p=2$$ the answer will be YES, but for $$p=11$$ the answer will be NO. Not sufficient.

(2) $$p$$ is a factor of $$\frac{(n+2)!}{n!}$$. Simplify: $$\frac{(n+2)!}{n!}=(n+1)(n+2)$$. Now, if $$n=2$$ then $$(n+1)(n+2)=12$$ and for $$p=2$$ the answer will be YES, but for $$p=3$$ the answer will be NO. Not sufficient.

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$. Now, $$(n+1)(n+2)-1$$ and $$(n+1)(n+2)$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example, 20 and 21 are consecutive integers, thus the only common factor they share is 1. So, as from (2) $$p$$ is a factor of $$(n+1)(n+2)$$ then it cannot be a factor of $$(n+1)(n+2)-1$$. Thus, in order for $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$. Sufficient.

Hi Bunuel,

Can n = 1?
_________________

If you like my post, please send some kudos! :D

Intern
Joined: 06 Feb 2016
Posts: 48
Location: Poland
Concentration: Finance, Accounting
GMAT 1: 730 Q49 V41
GPA: 3.5

### Show Tags

20 Sep 2016, 02:52
2
Aren't these questions unnecessarily difficult to be a GMAT question?
Intern
Joined: 10 Dec 2016
Posts: 24

### Show Tags

29 Dec 2016, 11:16
LOVE the question. But what are the chances of something like this showing up Bunuel? IT took me a good 10 mins to solve this :/
Intern
Joined: 24 Jun 2017
Posts: 6

### Show Tags

31 Oct 2017, 02:32
Can N=1??
Manager
Joined: 29 Feb 2016
Posts: 112
Location: India
GMAT 1: 700 Q49 V35
GPA: 3

### Show Tags

22 Nov 2017, 22:42
4
Bunuel wrote:
If $$n$$ is a positive integer and $$p$$ is a prime number, is $$p$$ a factor of $$n!$$?

(1) $$p$$ is a factor of $$(n+2)!-n!$$

(2) $$p$$ is a factor of $$\frac{(n+2)!}{n!}$$

I solved it in this way

stmt 1 : p is fact of ((n+2)(n+1)-1) * N!

P can be a factor of ((n+2)(n+1)-1) or P can be a factor of n!

Stmt 2 : P is afactor of (n+2)(n+1)*n!/n!

p is a factor of (n+2)(n+1)

Combined as p is a factor of (n+2)(n+1) , p cannot be factor of ((n+2)(n+1)-1).

So P is a factor of n!.

Can someone please validate this approach.
_________________

Find what you love and let it kill you. — Charles Bukowski

Math Expert
Joined: 02 Sep 2009
Posts: 52971

### Show Tags

22 Nov 2017, 23:06
sumanainampudi wrote:
Bunuel wrote:
If $$n$$ is a positive integer and $$p$$ is a prime number, is $$p$$ a factor of $$n!$$?

(1) $$p$$ is a factor of $$(n+2)!-n!$$

(2) $$p$$ is a factor of $$\frac{(n+2)!}{n!}$$

I solved it in this way

stmt 1 : p is fact of ((n+2)(n+1)-1) * N!

P can be a factor of ((n+2)(n+1)-1) or P can be a factor of n!

Stmt 2 : P is afactor of (n+2)(n+1)*n!/n!

p is a factor of (n+2)(n+1)

Combined as p is a factor of (n+2)(n+1) , p cannot be factor of ((n+2)(n+1)-1).

So P is a factor of n!.

Can someone please validate this approach.

_______________
That's correct.
_________________
Intern
Joined: 17 Apr 2018
Posts: 20
Location: India
GMAT 1: 710 Q49 V38
GPA: 4

### Show Tags

02 Aug 2018, 09:11
Bunuel
Great question !
I always wonder who you make these ques
Thread Master - Part Time MBA Programs
Joined: 11 Jan 2018
Posts: 137

### Show Tags

06 Sep 2018, 18:57
Bunuel wrote:
sumanainampudi wrote:
Bunuel wrote:
If $$n$$ is a positive integer and $$p$$ is a prime number, is $$p$$ a factor of $$n!$$?

(1) $$p$$ is a factor of $$(n+2)!-n!$$

(2) $$p$$ is a factor of $$\frac{(n+2)!}{n!}$$

I solved it in this way

stmt 1 : p is fact of ((n+2)(n+1)-1) * N!

P can be a factor of ((n+2)(n+1)-1) or P can be a factor of n!

Stmt 2 : P is afactor of (n+2)(n+1)*n!/n!

p is a factor of (n+2)(n+1)

Combined as p is a factor of (n+2)(n+1) , p cannot be factor of ((n+2)(n+1)-1).

So P is a factor of n!.

Can someone please validate this approach.

_______________
That's correct.

I get this approach, but shouldn't Statement 2 yield the answer as B if she got to "P is a factor of (n+2)(n+1), hence it's not a factor of n!?
Intern
Joined: 01 Aug 2018
Posts: 1

### Show Tags

10 Sep 2018, 06:42
1
For those of us who didn't catch the (n+1)(n+2) - 1 can't be a factor, there's another approach for combining statements:

Rearraging a bit, we have [(n+1)(n+2)(n!) - (n!)]/p = (n+1)(n+2)(n!)/ p - (n!) / p= an integer by (1).

By (2), we know that (n+1)(n+2)/p is an integer, so we have:

[(n+1)(n+2)(n!)/ p ]- [(n!) / p]= Integer * (n!) - n!/p = Integer. The only way that n!/p subtracted from an integer * n! can be an integer is if n!/p is an integer- which implies that n!/p = integer, meaning p is a factor of n!.
Re: M27-01   [#permalink] 10 Sep 2018, 06:42
Display posts from previous: Sort by

# M27-01

Moderators: chetan2u, Bunuel

 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®.