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The integers m and p are such that 2<m<p and m is not

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Director
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The integers m and p are such that 2<m<p and m is not [#permalink] New post 28 Feb 2008, 08:10
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?

(1)The greatest common factor of m and p is 2.
(2) The least common multiple of m and p is 30.
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Re: GMATprep problem [#permalink] New post 28 Feb 2008, 08:15
jimmyjamesdonkey 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?

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


1: seems to be suff. b/c both m and p must be even. and 2<m<p thus we can only have something like 6<8 8<10 etc...

thus r>1

2: 2<10<15 or 2<5<6. 15/10 r>1. 6/5 r=1 Insuff.


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Re: GMATprep problem [#permalink] New post 28 Feb 2008, 08:28
GMATBLACKBELT, The OA is A...and I had the same reasoning...but...

what about 2 < 4 < 8?

That doesn't work 8/4 = r0? What am I missing here?
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Re: GMATprep problem [#permalink] New post 28 Feb 2008, 08:28
Doh, nevermind...It's because m is not a factor of p correct?
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Re: GMATprep problem [#permalink] New post 28 Feb 2008, 11:56
Can someone explain how (2) The least common multiple of m and p is 30. is not suff?

I know it isn't...but would like to see how it is calculated. Thanks.
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Re: GMATprep problem [#permalink] New post 28 Feb 2008, 15:43
I just picked numbers 6/5 has a remainder 1 and a LCM of 30

then I tried to figure other numbers I was like 6*10 30 is LCM 10/6 R of 4
Re: GMATprep problem   [#permalink] 28 Feb 2008, 15:43
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