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

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Manager
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Integers m and p are such that 2<m<p and m in NOT a [#permalink] New post 20 Mar 2009, 12:49
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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.


Please explain your answers.
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Re: DS GMAT Prep - Integer [#permalink] New post 20 Mar 2009, 21:06
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
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Re: DS GMAT Prep - Integer [#permalink] New post 22 Mar 2009, 05:28
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.


Please explain your answers.
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Re: DS GMAT Prep - Integer [#permalink] New post 22 Mar 2009, 08:41
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, 08:41
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Integers m and p are such that 2<m<p and m in NOT a

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