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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????? _________________

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.

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 _________________

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.

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. _________________

learn the rules of the game, then play better than anyone else.

Re: The integers m and p are such that 2<m<p, and m is not a fac [#permalink]
09 Oct 2013, 08:16

3

This post received KUDOS

Expert's post

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.

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...