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

 It is currently 06 May 2015, 14:54

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

The integers m and p are such that 2 < m < p and m is

Author Message
TAGS:
Director
Joined: 03 Sep 2006
Posts: 885
Followers: 6

Kudos [?]: 259 [0], given: 33

The integers m and p are such that 2 < m < p and m is [#permalink]  17 Jun 2007, 18:03
00:00

Difficulty:

(N/A)

Question Stats:

100% (01:58) correct 0% (00:00) wrong based on 2 sessions
The integers m and p are such that 2 < m < p and m is not a factor of p. If r is the remainder when p is divided by m, is r > 1?

( i ) The greatest common factor of m and p is 2.

( ii ) the least common multiple of m and p is 30 .
VP
Joined: 10 Jun 2007
Posts: 1465
Followers: 6

Kudos [?]: 133 [0], given: 0

Re: DS- Integers m and p [#permalink]  18 Jun 2007, 19:38
LM wrote:
The integers m and p are such that 2 < m < p and m is not a factor of p. If r is the remainder when p is divided by m, is r > 1?

( i ) The greatest common factor of m and p is 2.

( ii ) the least common multiple of m and p is 30 .

I got A

(1) Greatest common factor of m & p is 2. We know that both p and m must be even. Given 2<m<p, the minimum difference between the two integers is 2. It is also given that m isn't a factor of p; thus, r must be greater than 1. SUFFICIENT.

(2) p = (30^x) * integer
m = (30^x) * integer
In this case, if you plug in the intergers, you can get anything. INSUFFICIENT.
Senior Manager
Joined: 06 Jul 2004
Posts: 474
Location: united states
Followers: 1

Kudos [?]: 17 [0], given: 0

Re: DS- Integers m and p [#permalink]  18 Jun 2007, 19:40
LM wrote:
The integers m and p are such that 2 < m <p> 1?

( i ) The greatest common factor of m and p is 2.

( ii ) the least common multiple of m and p is 30 .

Should be A.

m and p are both greater than 2 (according to the info).

consider (i)

m & p both are even numbers. the smallest of even numbers that satisfy (i) and the given information are 4 & 6. clearly remainder r > 1 for all cases.

consider (ii)

m=3, p=10
or
m=5, p=6
or
m=2, p=15

clearly, r is not greater than 1 in all cases.

please feel free to correct me.

Guys, it's not a poll, votes for A,B,C,D or E won't count. Please provide your answers with explanations so that we can have a healthy discussion.
_________________

for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

Intern
Joined: 03 Aug 2006
Posts: 23
Followers: 0

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

I got D because

(i) agree with explantions
(ii) The nos. can only be:
3,10 or 5,6 and not 2,15 as 2<m<p
for both 3,10 and 5,6 remainder is 1 hence less than 2. sufficient

I could be completely wrong for the second statement.
VP
Joined: 10 Jun 2007
Posts: 1465
Followers: 6

Kudos [?]: 133 [0], given: 0

ridhimagupta wrote:
I got D because

(i) agree with explantions
(ii) The nos. can only be:
3,10 or 5,6 and not 2,15 as 2<m<p
for both 3,10 and 5,6 remainder is 1 hence less than 2. sufficient

I could be completely wrong for the second statement.

Least common multiple isn't what you described.
Say you have 30 and 40.
30 = 2 *3 *5
40 = 2^3 *5

Lcm = 2^3 * 3 * 5 = 8 * 15 = 120.
Director
Joined: 14 Jan 2007
Posts: 780
Followers: 2

Kudos [?]: 71 [0], given: 0

I would go with 'D'.

Stmt1: the p/m could be 6/4, 12/10, 22/6....each of them has a remainder > 1
So SUFF

Stmt2: LCM is 30, p/m could be 6/5, 10/3, 15/2, 30/10...each of them has remainder <= 1
So SUFF
Intern
Joined: 03 Aug 2006
Posts: 23
Followers: 0

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

Stmt2: LCM is 30, p/m could be 6/5, 10/3, 15/2, 30/10...each of them has remainder <= 1
So SUFF

You meant 30/1 right????

This is what I am getting confused about - if the statement says 2<m<p then they have to positive integers greater than 2.
If the LCM is 30 then the only nos. which qualify with all these conditions have to be 5,6 and 3,10. In which case remainder is 1.

Am I missing something about LCMs here?
Senior Manager
Joined: 03 Jun 2007
Posts: 384
Followers: 2

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

I feel it is D. Agree with statement 1. But for 2 we can only have number 5,6 and 10,3 and in either case the remainder is 1 which is not greater than 1. So D
Manager
Joined: 28 Aug 2006
Posts: 160
Followers: 2

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

Agreed its D, Sine there are only 2 sets of numbers (3,10) and (5,6) that meet the coondition and their reminder is constant 1
Intern
Joined: 04 Feb 2007
Posts: 39
Followers: 1

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

'C' IT IS....

Stmt1: m= 4, 6, 8, 10
p= 6, 8, 10, 12......
so remainder(p/m) could be 6/4=2, 8/4=0, 10/4=2....
INSUFF

Stmt2: m= 3, 5, 6, 10, 3, 5, 6, 15
p = 10, 6, 10, 30, 30, 30, 30.......(PAIR WISE).........LCM is 30
SO p/m could be 6/5=1, 10/3=1, 10/6=4, 30/10=0...
So INSUFF

(1)+(2): p/m=10/6, 10/30, 6/30 fetching r=0 all the time which is less than 1 all through.SUFF

C
Director
Joined: 26 Feb 2006
Posts: 905
Followers: 4

Kudos [?]: 52 [0], given: 0

D is not necessarily correct. what if m = 6 and p = 10?

http://www.gmatclub.com/phpbb/viewtopic ... =remainder

vshaunak@gmail.com wrote:
I would go with 'D'.

Stmt1: the p/m could be 6/4, 12/10, 22/6....each of them has a remainder > 1
So SUFF

Stmt2: LCM is 30, p/m could be 6/5, 10/3, 15/2, 30/10...each of them has remainder <= 1
So SUFF
Director
Joined: 14 Jan 2007
Posts: 780
Followers: 2

Kudos [?]: 71 [0], given: 0

Himalayan wrote:
D is not necessarily correct. what if m = 6 and p = 10?

http://www.gmatclub.com/phpbb/viewtopic ... =remainder

vshaunak@gmail.com wrote:
I would go with 'D'.

Stmt1: the p/m could be 6/4, 12/10, 22/6....each of them has a remainder > 1
So SUFF

Stmt2: LCM is 30, p/m could be 6/5, 10/3, 15/2, 30/10...each of them has remainder <= 1
So SUFF

Thanks Himalayan for catching this. Yes 'D' shouldn't be right then.
Senior Manager
Joined: 11 Feb 2007
Posts: 359
Followers: 1

Kudos [?]: 41 [0], given: 0

A for me.

1) SUFF - agree with bkk145's explanation.
2) INSUFF consider m=10 p=15
Similar topics Replies Last post
Similar
Topics:
Integers m and p are such that 2<m<p and m in NOT a 3 20 Mar 2009, 12:49
1 The integers m and p are such that 2 < m < p and m is 1 12 Mar 2008, 09:57
4 The integers m and p are such that 2<m<p, and m is not a fac 13 16 Jun 2006, 17:05
The integers m and p are such that 2<m<p and m is not 7 06 Jan 2006, 08:05
The integers m and p are such that 2 < m < p and m is 6 26 Dec 2005, 17:16
Display posts from previous: Sort by