Factor p : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 21:12

### 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

# Factor p

Author Message
Manager
Joined: 28 Apr 2003
Posts: 93
Location: Singapore
Followers: 1

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

### Show Tags

08 Feb 2004, 07:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

257.

This is DS problem. Could anyone try it and show the workings?

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

1. k> 4!
2. 13! + 2 ≤ K ≤ 13! + 13
SVP
Joined: 30 Oct 2003
Posts: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 98 [0], given: 0

### Show Tags

08 Feb 2004, 08:23
A) k > 4! means k > 24 let K be 25 which has factors 1, 5 and 25
1 < 5 < 25
But 29 has factors 1 and 29 and does not satisfy the given condition
So A is insufficient.
B) 13! has all the integer mutiples from 1 and 13
13!+2 is divisible by 2
13!+3 is divisible by 3 and so on
13!+12 is divisible by 12
For all of these 1 < n(from 2 to 12) < 13!+n

B is suffiient.
Senior Manager
Joined: 30 Aug 2003
Posts: 324
Location: dallas , tx
Followers: 1

Kudos [?]: 26 [0], given: 0

### Show Tags

08 Feb 2004, 16:10
anand.. could u pls.. tell me how to crack the sencond sentence.. is any easy or diff way.. what u did is good but a bit confusing..
_________________

shubhangi

SVP
Joined: 30 Oct 2003
Posts: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 98 [0], given: 0

### Show Tags

08 Feb 2004, 19:03
I wont be surprised if the answer is D. The stem of the problem is confusing. But I am pretty sure that B is the answer.

1) k>4! involves infinite numbers. This set holds primes and non primes.
Only primes do not satisfy the condition 1<p<k
2) Now 13! = 13*12*11*10*9*8*7*6*5*4*3*2*1 = X
This huge number is divisible by any integer between 1 and 13.
The second condition says that the number K is in between
13!+2 and 13!+13. So what are the possible numbers

13*12*11*10*9*8*7*6*5*4*3*2*1 + 3 divisible by p=3 1<3<(X+3)
13*12*11*10*9*8*7*6*5*4*3*2*1 + 4 divisible by p=4 1<4<(X+4)
----
----
13*12*11*10*9*8*7*6*5*4*3*2*1 + 12 divisible by p=3 1<12<(X+12)

So any number you take it has a factor > 1 and < that number.
Hence B is sufficient

I hope I made it clear.
08 Feb 2004, 19:03
Display posts from previous: Sort by
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}