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# The integers m and p are such that 2 is less than m and m is

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VP
Joined: 30 Jun 2008
Posts: 1026
The integers m and p are such that 2 is less than m and m is [#permalink]

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02 Nov 2008, 07:17
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33% (00:48) correct 67% (01:47) wrong based on 6 sessions

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The integers m and p are such that 2 is less than m and m is less than p. Also, m is not a factor of p. If r is the remainder when p is divided by m, is r > 1.

1. The greatest common factor of m and p is 2.
2. The least common multiple of m and p is 30.
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Senior Manager
Joined: 21 Apr 2008
Posts: 269
Location: Motortown
Re: DS : GMATPrep : LCM GCD [#permalink]

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02 Nov 2008, 07:40
A

2<m<p
p=mx+r x=0,1,2,3,....

stmt1: GCD of m and p is 2
So, m and p can be 4,6 - r>1 - Suff

Stmt2: LCM of m and p is 30
m and p can be
5,6 - r=1
3,10 - r>1 - InSuff
VP
Joined: 30 Jun 2008
Posts: 1026
Re: DS : GMATPrep : LCM GCD [#permalink]

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02 Nov 2008, 07:48
LiveStronger wrote:
A

2<m<p
p=mx+r x=0,1,2,3,....

stmt1: GCD of m and p is 2
So, m and p can be 4,6 - r>1 - Suff

Stmt2: LCM of m and p is 30
m and p can be
5,6 - r=1
3,10 - r>1 - InSuff

for statement 1, what if m=6 and p=10, GCD is still 2. and remainder is 4 ?? or 4/6 is considered as 2/3 and hence remainder 2. This is what confused me ...
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Senior Manager
Joined: 21 Apr 2008
Posts: 269
Location: Motortown
Re: DS : GMATPrep : LCM GCD [#permalink]

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02 Nov 2008, 07:53
even when you consider 4 and 10, r >1

You can't consider 4/6 as 2/3, because
GCD for 2 and 3 is not 2
m and p should be >2

What is the OA ?
VP
Joined: 30 Jun 2008
Posts: 1026
Re: DS : GMATPrep : LCM GCD [#permalink]

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02 Nov 2008, 07:58
LiveStronger wrote:
even when you consider 4 and 10, r >1

You can't consider 4/6 as 2/3, because
GCD for 2 and 3 is not 2
m and p should be >2

What is the OA ?

OA is A. I get it now .. I was making a silly mistake
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

SVP
Joined: 17 Jun 2008
Posts: 1529
Re: DS : GMATPrep : LCM GCD [#permalink]

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02 Nov 2008, 23:14
Another way of looking at the problem is is the difference between p and multiple of m >1, or in simplest form, is the difference between p and m > 1.

From stmt1: Since, 2 is a common factor between m and p and p>m>2, hence p-m will at least be 2. Hence, sufficient.
From stmt2: m and p can be 5 and 6 and difference = 1. However, if m=3 and p = 10, difference >1. Hence, insufficient.
Re: DS : GMATPrep : LCM GCD   [#permalink] 02 Nov 2008, 23:14
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