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The integers m and p are such that 2 < m < p and m is

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The integers m and p are such that 2 < m < p and m is [#permalink] New post 26 Dec 2005, 17:16
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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: DS: Remainder [#permalink] New post 26 Dec 2005, 17:32
TeHCM 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

A.
from i, m and p are multiples of 2 but m and p canot be 4 and 8 or so on since 4 is a factor of 8. so reminder r is atleast 2 or more. suff

from ii, m and p can be 5 and 6 or 10 or 15 respectively. in first case, r =1 and in second case r =5. so not suff...
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Re: DS: Remainder [#permalink] New post 26 Dec 2005, 17:36
HIMALAYA wrote:
TeHCM 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

A.
from i, m and p are multiples of 2 but m and p canot be 4 and 8 or so on since 4 is a factor of 8. so reminder r is atleast 2 or more. suff

from ii, m and p can be 5 and 6 or 10 or 15 respectively. in first case, r =1 and in second case r =5. so not suff...


I don't know why but I was confused as to what a remainder was :roll:
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Re: DS: Remainder [#permalink] New post 26 Dec 2005, 18:24
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from ii, m and p can be 5 and 6 or 10 or 15 respectively. in first case, r =1 and in second case r =5. so not suff...


m and p respectively can be 3, 10 or 5, 6 with remiander 1 or 10, 15 with remainder 5. Only A.
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Re: DS: Remainder [#permalink] New post 26 Dec 2005, 18:36
TeHCM 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



Since p and m are integers, r must range from 0 to (m-1 ) ...because p is not divisible by m --> r can't be 0 ...now we have to consider the case of r=1

(1) from the statement ----> both m and p are even
if r=1 ----> p = m*k+1 (k is integer) . But m is even ---> m*k +1 is odd---> unreasonable coz p must be even ----> r CAN'T be 1 ---> r >1
--->suff.

(2) the same to others

A it is.
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 [#permalink] New post 27 Dec 2005, 00:40
Good work as always

OA is A
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Re: DS: Remainder [#permalink] New post 28 Dec 2005, 04:24
TeHCM 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


Pretty late to see this question, but better late than never.

From stmt1 m and p are even , and since m#p, the remainder r will be greater than 1.

For ex. Let m = 4 and p = 6 then r=2.
Let m = 8 and p = 10 then r = 2
So stmt 1 is suff.

From stmt 2 m could be 5 and p could be 6. or m could be 2 and p could be 15. So insuff.

Hence A.
Re: DS: Remainder   [#permalink] 28 Dec 2005, 04:24
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The integers m and p are such that 2 < m < p and m is

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