Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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]

Show Tags

09 Oct 2013, 09:16

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.

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...