Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 26 May 2017, 12:07

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

Integers m and p are such that 2<m<p and m in NOT a

Author Message
Manager
Joined: 19 Oct 2008
Posts: 94
Followers: 1

Kudos [?]: 45 [0], given: 0

Integers m and p are such that 2<m<p and m in NOT a [#permalink]

Show Tags

20 Mar 2009, 13:49
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

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

Integers m and p are such that 2<m<p and m in NOT a factor of p. If r is the remainder when p is divided by m, is r>1?

1. Greatest common factor of m and p is 2.
2. Least common multiple of m and p is 30.

Senior Manager
Joined: 08 Jan 2009
Posts: 327
Followers: 2

Kudos [?]: 159 [0], given: 5

Re: DS GMAT Prep - Integer [#permalink]

Show Tags

20 Mar 2009, 22:06
Given:

2<m<p . m is not factor of p.

Stmt 1 :
m = 2x p = 2y

so u know m and p are multiples of 2 . So if you take different values
m = 4 p = 6 then r = 2
m=8 p =10 then r =2
so defintely r >`1. Sufficient

Stmt 2 :

m and p will divide mutiples of 30.

m= 5 p = 6 then r =1
m=6 p =10 then r = 4

so r can be less or greater than 1. Insufficient

Ans. A
Director
Joined: 01 Apr 2008
Posts: 882
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 30

Kudos [?]: 705 [0], given: 18

Re: DS GMAT Prep - Integer [#permalink]

Show Tags

22 Mar 2009, 06:28
My explanation is somewat different:)

Given, p = mA + r ( m is some integer)
1)
m and p have 2 as the greatest common factor, so
m = 2 * I1
p = 2 * I2

now, 2*I2 = 2*I1 + r , but it is given that , p > m , so, I2 > I1 => I2-I1 > 0

therefore, 2 (I2-I1) = r => r > 1.

2)
m and p have 30 as the LCM.
m = 2*3*5 * I1 ( but I1 can have more 2's 3's or 5's).
p = 2*3*5 * I2 ( but I2 can have more 2's 3's or 5's).

So we cant have a relation between I2 and I1. hence we cant prove that r > 1.

Accountant wrote:
Integers m and p are such that 2<m<p and m in NOT a factor of p. If r is the remainder when p is divided by m, is r>1?

1. Greatest common factor of m and p is 2.
2. Least common multiple of m and p is 30.

Intern
Joined: 07 Oct 2008
Posts: 12
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: DS GMAT Prep - Integer [#permalink]

Show Tags

22 Mar 2009, 09:41
Using 1 GCF =2
you can have (p,m) as (2*n, 2*(n+1)) where n = 2,3...
all have remainder = 2

Using 2 LCM =30 , I plugged some num combination
(p,m) can have (6,5) remainder = 1
and (30,15) remainder = 0

so it is A
Re: DS GMAT Prep - Integer   [#permalink] 22 Mar 2009, 09:41
Display posts from previous: Sort by