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

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The integers m and p are such that 2 < m <p. m is not [#permalink] New post 08 Jun 2007, 10:00
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The integers m and p are such that 2 < m <p.
m is not a factor of p. r is the remainder when p is divided by m.
Q: Is r more than 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: DS - remainder GMATPrep [#permalink] New post 08 Jun 2007, 23:29
Caas wrote:
The integers m and p are such that 2 < m <p.
m is not a factor of p. r is the remainder when p is divided by m.

Q: Is r more than 1?

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


1. if the gratest common factor of m and p is 2, then m and p, both, are even. m and p can not be 4 and 8 cuz in that case the common factor is 4 not 2. therefore in that case r is obviously grater than 1. suff.

2. if the least common multiple of m and p is 30, m and p could be 5 and 6 respectively ot they could be 3 or 10 respectively or 15 and 30. so not suff.

So A.
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 [#permalink] New post 09 Jun 2007, 00:02
D.

Both are sufficient.

S1. m and n are 4,6 or 6,8 or 8,10 or 10,12 or 12,14
r>1
SUFF

S2. Only 2 combinations here.
5*6
3*10
Both r=1
SUFF
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 [#permalink] New post 09 Jun 2007, 00:06
not necessarily is D correct. if m = 6 and p = 10.
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 [#permalink] New post 09 Jun 2007, 00:10
Youre right Himalayan :shock:
  [#permalink] 09 Jun 2007, 00:10
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