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

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The integers m and p are such that 2<m<p and m is not [#permalink]  16 Jun 2006, 18:05
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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?

condition 1: the greatest common factor of m and p is 2.

condition 2: the least common multiple of m and p is 30.
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Either there is a typo or I am a jerk

Following is the given information:

1. m and p are integers
2. 2<m<p
3. m is not a factor of p.

If r is the remainder when p is divided by m, is r>1? I think r will always be greater than 1. Because

1. p is atleast 1 greater than m because of conditions 1 and 2 above.
2. r can not be zero because of condition 3 above.
3. a remainder is never a real number.

So we don't need to read the statements to answer the question?????
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ps_dahiya wrote:
Either there is a typo or I am a jerk

Following is the given information:

1. m and p are integers
2. 2<m<p
3. m is not a factor of p.

If r is the remainder when p is divided by m, is r>1? I think r will always be greater than 1. Because

1. p is atleast 1 greater than m because of conditions 1 and 2 above.
2. r can not be zero because of condition 3 above.
3. a remainder is never a real number.

So we don't need to read the statements to answer the question?????

consider 2<3<4

4/3--> remainder = 1 which is not greater than 1.

r will not always be greater.

I would go with A here.

St1: GCD is 2 so that means they are both even and will differ by at least 2. So Suff.

St2.: Stands for 10 and 3 but fails for 5 and 6 so Insuff.

Hence A
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Re: DS: Number Properties (11) [#permalink]  16 Jun 2006, 22:04
mrmikec 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?

condition 1: the greatest common factor of m and p is 2.
condition 2: the least common multiple of m and p is 30.

do not see anything wrong with the question.

go with A.
from 1, the clue is p>m>2. so none of p and m can be 2. if so, the r is >1.

in 2, p and m could be 6 and 5 or 10 and 3 respectively. in both cases, r =1. not suff..
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tapan22 wrote:
ps_dahiya wrote:
Either there is a typo or I am a jerk

Following is the given information:

1. m and p are integers
2. 2<m<p
3. m is not a factor of p.

If r is the remainder when p is divided by m, is r>1? I think r will always be greater than 1. Because

1. p is atleast 1 greater than m because of conditions 1 and 2 above.
2. r can not be zero because of condition 3 above.
3. a remainder is never a real number.

So we don't need to read the statements to answer the question?????

consider 2<3<4

4/3--> remainder = 1 which is not greater than 1.

r will not always be greater.

I would go with A here.

St1: GCD is 2 so that means they are both even and will differ by at least 2. So Suff.

St2.: Stands for 10 and 3 but fails for 5 and 6 so Insuff.

Hence A

Agree. I don't know what I was thinking
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No worries psdahiya, I have seen your posts and I know you are good. I guess it was just really late that you were working on this problem.

Hang in there
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why stmt2 is not valid ? it gives a answer no.

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gmatinjune wrote:
why stmt2 is not valid ? it gives a answer no.

\

yes it should be D

1. the reminder is always > 1 - YES
2. the remnder is always 1 - NO

whats the OA?
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tapan22 wrote:
ps_dahiya wrote:
Either there is a typo or I am a jerk

Following is the given information:

1. m and p are integers
2. 2<m<p
3. m is not a factor of p.

If r is the remainder when p is divided by m, is r>1? I think r will always be greater than 1. Because

1. p is atleast 1 greater than m because of conditions 1 and 2 above.
2. r can not be zero because of condition 3 above.
3. a remainder is never a real number.

So we don't need to read the statements to answer the question?????

consider 2<3<4

4/3--> remainder = 1 which is not greater than 1.

r will not always be greater.

I would go with A here.

St1: GCD is 2 so that means they are both even and will differ by at least 2. So Suff.

St2.: Stands for 10 and 3 but fails for 5 and 6 so Insuff.

Hence A

Just to verify my thought process...

So C1. states that the GCD will also be even but with a multiple that is both odd and not factors of each other. (the oddness takes care of the C1 because every even number is divisible by 2 and since they are multiples that are not factors of each other eliminates the possibility that they are factors.)

So for example

6 (2*3) and 10 (2*5) 10/6 R=4
10 (2*5) and 12 (2*2*3) 12/10 R=2

So R>1 because you are dividing two different odd multiples that are not factors of each other (3,5,7,11,13), hence the difference will always be at least 2.

C2. gives 5,6 ---> 6/5 R=1
or 15/6 15/6 ---> R=3
INSUFFICIENT
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That is correct Mrmikec.

I still think it is A. I will just wait for the OA.
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tapan22 wrote:
That is correct Mrmikec.

I still think it is A. I will just wait for the OA.

Excuse my oversight. tHE OA is A.
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tapan22 wrote:
No worries psdahiya, I have seen your posts and I know you are good. I guess it was just really late that you were working on this problem.

Hang in there

Thanks for the encouragement buddy. I badly need that
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# The integers m and p are such that 2<m<p and m is not

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