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The integers m and p are such that 2<m<p, and m is not a fac [#permalink]
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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? (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: theintegersmandparesuchthat2mpandmisnota101360.html
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Last edited by Bunuel on 09 Oct 2013, 09:15, edited 1 time in total.
Edited the question and added the OA.



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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]
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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.
shouldnt the answer be D



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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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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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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]
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09 Oct 2013, 09:16
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: theintegersmandparesuchthat2mpandmisnota101360.html
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Re: The integers m and p are such that 2<m<p, and m is not a fac
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