Find all School-related info fast with the new School-Specific MBA Forum

It is currently 18 Sep 2014, 12:23

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Does the integer K have a factor P such that 1<P<K? 1)

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
avatar
Joined: 02 Feb 2004
Posts: 345
Followers: 1

Kudos [?]: 10 [0], given: 0

GMAT Tests User
Does the integer K have a factor P such that 1<P<K? 1) [#permalink] New post 26 May 2004, 15:54
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Does the integer K have a factor P such that 1<P<K?
1) K>4!
2)13!+2<=K<=13!+13

please show & explain workings :lol:
SVP
SVP
User avatar
Joined: 30 Oct 2003
Posts: 1798
Location: NewJersey USA
Followers: 4

Kudos [?]: 34 [0], given: 0

GMAT Tests User
 [#permalink] New post 26 May 2004, 19:26
A) K>4! K = { 25,26,27,28,29...} 29 is a prime and has no factor p such that 1<p<k
But 25 and 26 do So Insufficient

2)
13!+2 is divisible by 2 hence 1<2<(13!+2)
13!+3 is divisible by 3 hence 1<3<(13!+3)
13!+4 is divisible by 4 hence 1<4<(13!+4)
13!+5 is divisible by 5 hence 1<5<(13!+5)
13!+6 is divisible by 6 hence 1<6<(13!+6)
13!+7 is divisible by 7 hence 1<7<(13!+7)
13!+8 is divisible by 8 hence 1<8<(13!+8)
13!+9 is divisible by 9 hence 1<9<(13!+9)
13!+10 is divisible by 10 hence 1<10<(13!+10)
13!+11 is divisible by 11 hence 1<11<(13!+11)
13!+12 is divisible by 12 hence 1<12<(13!+12)
13!+13 is divisible by 13 hence 1<12<(13!+13)

All the integers between 13!+2 and 13!+13 have one factor p
such that 1<p<k
Sufficient.
Senior Manager
Senior Manager
avatar
Joined: 02 Feb 2004
Posts: 345
Followers: 1

Kudos [?]: 10 [0], given: 0

GMAT Tests User
 [#permalink] New post 27 May 2004, 06:29
anandnk wrote:
A) K>4! K = { 25,26,27,28,29...} 29 is a prime and has no factor p such that 1<p<k
But 25 and 26 do So Insufficient

2)
13!+2 is divisible by 2 hence 1<2<(13!+2)
13!+3 is divisible by 3 hence 1<3<(13!+3)
13!+4 is divisible by 4 hence 1<4<(13!+4)
13!+5 is divisible by 5 hence 1<5<(13!+5)
13!+6 is divisible by 6 hence 1<6<(13!+6)
13!+7 is divisible by 7 hence 1<7<(13!+7)
13!+8 is divisible by 8 hence 1<8<(13!+8)
13!+9 is divisible by 9 hence 1<9<(13!+9)
13!+10 is divisible by 10 hence 1<10<(13!+10)
13!+11 is divisible by 11 hence 1<11<(13!+11)
13!+12 is divisible by 12 hence 1<12<(13!+12)
13!+13 is divisible by 13 hence 1<12<(13!+13)

All the integers between 13!+2 and 13!+13 have one factor p
such that 1<p<k
Sufficient.


wow! I admire your patience. But anyone with a shortcut method?
Intern
Intern
avatar
Joined: 04 May 2004
Posts: 6
Location: Midwest
Followers: 0

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

hi [#permalink] New post 27 May 2004, 08:08
Hi Anand,

Is there a general number property like

n!+k is divisible by k or did you do actually figure it out for each one of these numbers?

Thanks
SVP
SVP
User avatar
Joined: 30 Oct 2003
Posts: 1798
Location: NewJersey USA
Followers: 4

Kudos [?]: 34 [0], given: 0

GMAT Tests User
 [#permalink] New post 27 May 2004, 09:21
Hi,

No formula for this. I just did it in my brain but I wanted to explain in detail so I wrote so much. Actually kpadma asked me to solve the same problem once during yahoo chat. I remembered the explanation I gave and just reproduced it here.

any number x!+y is divisible by y as long as y <= x because y is a factor of x
if y > x then cannot say for sure.

Anand.
Intern
Intern
avatar
Joined: 04 May 2004
Posts: 6
Location: Midwest
Followers: 0

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

Thanks [#permalink] New post 27 May 2004, 11:11
Hi Anand,

Thank you for your explanation, it makes sense now.
Manager
Manager
avatar
Joined: 30 Apr 2004
Posts: 50
Location: NYC
Followers: 1

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

Re: DS:inequalities :challanging [#permalink] New post 28 May 2004, 08:44
mirhaque wrote:
Does the integer K have a factor P such that 1<P<K?
1) K>4!
2)13!+2<=K<=13!+13

please show & explain workings :lol:


The question can be rephrased as is K prime, because if K being prime is the only time that 1<P<K will not hold true.


(1) is very insufficient. It just says that K>24. Plenty of primes and non-primes out there.

(2) You gotta find out if there are any primes btwn 13!+2 and 13!+13

If you add a number to a multiple of that number, the sum will also be a multiple of the number (i.e. 4 is a multiple of 2, so 4+2 will also be a multiple of 2). Which means that the sum cannot be prime.

13! is a multiple of every # btwn 1 and 13, inclusive. Therefore, if you add any number btwn 1 and 13 to 13!, the sum will be a multple of the addend and thus will not be prime. Hence (2) is sufficient.

B is the answer.
Manager
Manager
User avatar
Joined: 07 May 2004
Posts: 184
Location: Ukraine, Russia(part-time)
Followers: 2

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

GMAT Tests User
Re: DS:inequalities :challanging [#permalink] New post 29 May 2004, 02:32
mirhaque wrote:
Does the integer K have a factor P such that 1<P<K?
1) K>4!
2)13!+2<=K<=13!+13

please show & explain workings :lol:


1 is not sufficient: 25 and 29 are the examples: 25 = 5*5, 29 is prime number.

2 is sufficient to answer the question: for every 2 <= m <= 13, 13! + m :: m.
Re: DS:inequalities :challanging   [#permalink] 29 May 2004, 02:32
    Similar topics Author Replies Last post
Similar
Topics:
2 Does the integer K have a factor p such that 1<p<k? 1. vksunder 5 13 Jun 2008, 10:23
Experts publish their posts in the topic Does the integer k have factor p such that 1<p<k? 1) puma 4 16 Apr 2008, 08:58
Does integer K have a factor P such that 1<P<K ? 1. blog 3 31 Jan 2008, 19:57
Does the integer k have a factor p such that 1<p<k? 1) netcaesar 4 15 Oct 2006, 12:17
Does the integer k have a factor p such that 1<p<k? 1. rabbithole 2 24 Apr 2006, 07:40
Display posts from previous: Sort by

Does the integer K have a factor P such that 1<P<K? 1)

  Question banks Downloads My Bookmarks Reviews Important topics  


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

Powered by phpBB © phpBB Group and phpBB SEO

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®.