Does the integer k have a factor p such that 1 < p < : DS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 25 Feb 2017, 11:59

### 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

# Does the integer k have a factor p such that 1 < p <

Author Message
Intern
Joined: 03 Jul 2009
Posts: 15
Followers: 0

Kudos [?]: 39 [0], given: 5

Does the integer k have a factor p such that 1 < p < [#permalink]

### Show Tags

04 Oct 2009, 10:13
00:00

Difficulty:

(N/A)

Question Stats:

50% (00:00) correct 50% (00:18) wrong based on 2 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Does the integer k have a factor p such that 1 < p < k?

(1) k > 4!

(2) 13! +2 $$<=$$ k $$<=$$ 13! + 13

Can someone help me understand how to solve this problem?
[Reveal] Spoiler:
Senior Manager
Joined: 18 Aug 2009
Posts: 328
Followers: 8

Kudos [?]: 304 [1] , given: 13

Re: Number Properties and Inequalities Question [#permalink]

### Show Tags

04 Oct 2009, 10:26
1
KUDOS
Navigator wrote:
Does the integer k have a factor p such that 1 < p < k? - Question is simply asking if K is a PRIME number or not.

(1) k > 4! - It is not possible to find if K is a prime number. Insufficient.

(2) 13! +2 $$<=$$ k $$<=$$ 13! + 13 -
13! has factors {2, 3, 5, 7, 11, 13}.
So (13!+2) is divisible by 2 and (13!+13) is divisible by 13. Also, any number between (13!+2) and (13!+13) must be divisible by one of the factors of 13!. So this is sufficient.

Can someone help me understand how to solve this problem?
[Reveal] Spoiler:

___________________________-
Consider KUDOS for good posts
Intern
Joined: 03 Jul 2009
Posts: 15
Followers: 0

Kudos [?]: 39 [0], given: 5

Re: Number Properties and Inequalities Question [#permalink]

### Show Tags

04 Oct 2009, 14:06
Thank you! I understand the logic you applied to the two statements, but what was it about the question stem that lets you know that the question was ultimately asking if K is prime?
Senior Manager
Joined: 18 Aug 2009
Posts: 328
Followers: 8

Kudos [?]: 304 [0], given: 13

Re: Number Properties and Inequalities Question [#permalink]

### Show Tags

04 Oct 2009, 15:53
Well, the question is: Does the integer k have a factor p such that 1 < p < k. That means is there a factor of k other than 1 and k itself? It indicates that k is a prime factor if there is no other factor than 1 and k itself.

Hope it is clear now.
Intern
Joined: 03 Jul 2009
Posts: 15
Followers: 0

Kudos [?]: 39 [0], given: 5

Re: Number Properties and Inequalities Question [#permalink]

### Show Tags

05 Oct 2009, 04:53
Much clearer now, thanks again.
Intern
Joined: 08 Jul 2009
Posts: 28
Followers: 0

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

Re: Number Properties and Inequalities Question [#permalink]

### Show Tags

11 Oct 2009, 04:07
hgp2k wrote:
Navigator wrote:
Does the integer k have a factor p such that 1 < p < k? - Question is simply asking if K is a PRIME number or not.

(1) k > 4! - It is not possible to find if K is a prime number. Insufficient.

(2) 13! +2 $$<=$$ k $$<=$$ 13! + 13 -
13! has factors {2, 3, 5, 7, 11, 13}.
So (13!+2) is divisible by 2 and (13!+13) is divisible by 13. Also, any number between (13!+2) and (13!+13) must be divisible by one of the factors of 13!. So this is sufficient.

Can someone help me understand how to solve this problem?
[Reveal] Spoiler:

___________________________-
Consider KUDOS for good posts

Can you tell me please how did we know that any number between 13!+2 and 13!+13 must be divisble by one of the factors of 13! and is that applicable on all numbers with !
I mean i tried it with 3! and found that 3!+5 is for example prime number..

Director
Joined: 01 Apr 2008
Posts: 897
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 28

Kudos [?]: 658 [0], given: 18

Re: Number Properties and Inequalities Question [#permalink]

### Show Tags

11 Oct 2009, 06:36
rshawabka,
hgp2k has given a good explanation, for more details you can refer
http://www.beatthegmat.com/ds-question-t1514.html

The rule is : In a sum of two numbers, if those two numbers have a common factor then the sum will also have that factor ( i.e. the sum will be divisible by the common factor )
Manager
Joined: 05 Jul 2009
Posts: 182
Followers: 1

Kudos [?]: 49 [0], given: 5

Re: Number Properties and Inequalities Question [#permalink]

### Show Tags

23 Oct 2009, 08:57
Thanks economist.....
Re: Number Properties and Inequalities Question   [#permalink] 23 Oct 2009, 08:57
Display posts from previous: Sort by