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

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The integers m and p are such that 2<m<p, and m is not a fac [#permalink] New post 16 Jun 2006, 17:05
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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?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-integers-m-and-p-are-such-that-2-m-p-and-m-is-not-a-101360.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Oct 2013, 08:15, edited 1 time in total.
Edited the question and added the OA.
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 [#permalink] New post 16 Jun 2006, 18:12
Either there is a typo or I am a jerk :shock: :shock: :shock:

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????? :roll: :roll: :roll: :roll:
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 [#permalink] New post 16 Jun 2006, 19:09
ps_dahiya wrote:
Either there is a typo or I am a jerk :shock: :shock: :shock:

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????? :roll: :roll: :roll: :roll:


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] New post 16 Jun 2006, 21:04
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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 [#permalink] New post 16 Jun 2006, 22:17
tapan22 wrote:
ps_dahiya wrote:
Either there is a typo or I am a jerk :shock: :shock: :shock:

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????? :roll: :roll: :roll: :roll:


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 :x :x :x
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 [#permalink] New post 17 Jun 2006, 06:56
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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 [#permalink] New post 17 Jun 2006, 07:28
why stmt2 is not valid ? it gives a answer no.

shouldnt the answer be D
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 [#permalink] New post 17 Jun 2006, 08:50
gmatinjune wrote:
why stmt2 is not valid ? it gives a answer no.

shouldnt the answer be D
\


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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 [#permalink] New post 17 Jun 2006, 10:57
tapan22 wrote:
ps_dahiya wrote:
Either there is a typo or I am a jerk :shock: :shock: :shock:

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????? :roll: :roll: :roll: :roll:


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
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 [#permalink] New post 17 Jun 2006, 11:13
That is correct Mrmikec.

I still think it is A. I will just wait for the OA.
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 [#permalink] New post 17 Jun 2006, 11:15
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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 [#permalink] New post 17 Jun 2006, 19:49
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 :wink: :wink:
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Re: [#permalink] New post 09 Oct 2013, 08:00
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Okay, let's see what happens here:

we are given that:
2<m<p
p is not a multiple of m
they are both integers.

we want to assess if p/m yields a R>1.

st.1

Since the GCF of two numbers is basically the highest number that evenly divides both numbers, then m and p will be multiples of two (p must not be a multiple of m don't consider those cases). Let's just pick a some numbers and see how the outcome behaves, remember that 2<m<p which is a pivotal detail.

if m=4 and p=6 then R=2
if m=6 and p=16 then R=4
if m=8 and p=10 then R=2

we could test a slew of numbers and always obtain the same outcome: R>1. Suff.

st.2

Here we are coping with the LCM, and the LCM between two numbers is the product between all the different factors of both numbers with the highest power.
LCM of m and p = 30
LCM= 5(2)(3) we don't really know how many factors belong to m rather than to p, and we even do not care about, let's just test a couple of cases.

if m=9 and p=10 then R=1
if m=10 and p=27 then R=7 Non suff.

and the answer is A.
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Re: The integers m and p are such that 2<m<p, and m is not a fac [#permalink] New post 09 Oct 2013, 08:16
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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?

Given: 2<m<p and \frac{p}{m}\neq{integer}, which means that r>0. Note here that as 0<2<m<p then your example of -8 and 6 is not valid as both m and p are positive.
Question: r=?

(1) the greatest common factor of m and p is 2 --> both p and m are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since remainder is not zero (as \frac{p}{m}\neq{integer}), then remainder must be more than 1: 2, 4, ... Sufficient.

(2) the least common multiple of m and p is 30 --> if m=5 and p=6, remainder=1=1, answer to the question would be NO. BUT if m=10 and p=15 remainder=5>1 answer to the question would be YES. Two different answers. Not sufficient.

Answer: A.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-integers-m-and-p-are-such-that-2-m-p-and-m-is-not-a-101360.html
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Re: The integers m and p are such that 2<m<p, and m is not a fac   [#permalink] 09 Oct 2013, 08:16
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