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PREP -- D2

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PREP -- D2 [#permalink]

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New post 11 Sep 2008, 21:37
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pls explain the meaning of this problem to me...
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New post 11 Sep 2008, 21:55
//A//
1: so m and p both are even and m is not a factor of p, r if p is divided by m will be at least 2. so suff..
2: if m = 5 and p = 6, r is 1. no.
if m = 10 and p = 15, r is 5. yes.
so nsf..
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Re: PREP -- D2 [#permalink]

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New post 13 Sep 2008, 01:35
since both m and p are integers, should'nt r always be greater than or equal to 1..

for ex 5 divided by 4 or 8 divided by 3...

can someone please come up with a possibility where r is less than 1 except when there is perfect divisibility?
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Re: PREP -- D2 [#permalink]

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New post 13 Sep 2008, 09:06
In addition to the question above,


so m and p both are even

Is this always true when the greatest common factor is 2? Is there a specific pattern for this?


GMAT TIGER wrote:
//A//
1: so m and p both are even and m is not a factor of p, r if p is divided by m will be at least 2. so suff..
2: if m = 5 and p = 6, r is 1. no.
if m = 10 and p = 15, r is 5. yes.
so nsf..

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Re: PREP -- D2 [#permalink]

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New post 14 Sep 2008, 01:39
bigfernhead wrote:
In addition to the question above,


so m and p both are even

Is this always true when the greatest common factor is 2? Is there a specific pattern for this?


GMAT TIGER wrote:
//A//
1: so m and p both are even and m is not a factor of p, r if p is divided by m will be at least 2. so suff..
2: if m = 5 and p = 6, r is 1. no.
if m = 10 and p = 15, r is 5. yes.
so nsf..


1)2*4 =p ,2*3=m in this case r=1
2)LCM=30=5*3*2 =>2,3,5,6,10,15,30 are the possible numbers INSUFFI

(1) and (2) SUFFI remainder!=1 henceANSWERS IMO C
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Re: PREP -- D2 [#permalink]

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New post 14 Sep 2008, 02:31
arjtryarjtry wrote:
pls explain the meaning of this problem to me...

1. 2<m<mx+r

If they are both divisible by 2. m and mx+r are both even integers.
m is even integer ==> mx is even integer ==> r are even integer.
Thus mean r>1. ==> SUFF
2. Others mentioned. INSUFF.

A is correct answer.

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Re: PREP -- D2   [#permalink] 14 Sep 2008, 02:31
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