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:

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

04 Dec 2010, 08:58

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Fijisurf wrote:

I guess I am still not sure why in the original post sandipchowdhury wrote:

sandipchowdhury wrote:

St 1: when n is divided by 21 ( 7 and 3) the remainder is an odd number. But it cannot be 7, 3 or 9 . Hence the possibilities are : 1 and 5.

Why remainder cannot be 7, 3 or 9? (because in case of 28 divided by 21 remainder is 7)

OK. It's just not correct.

When n is divided by 21 the remainder is an odd number --> \(n=21q+odd=7*3q+odd\). Now, this odd number can be ANY odd number from 1 to 19, inclusive.

As for r: If \(n=22\) then \(n\) divided by 21 gives remainder of 1 and \(n\) divded by 7 also gives remainder of 1; If \(n=24\) then \(n\) divided by 21 gives remainder of 3 and \(n\) divded by 7 also gives remainder of 3; If \(n=26\) then \(n\) divided by 21 gives remainder of 5 and \(n\) divded by 7 also gives remainder of 5; If \(n=28\) then \(n\) divided by 21 gives remainder of 7 and \(n\) divded by 7 gives remainder of 0.

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

13 Dec 2010, 20:55

Great thanks guys. Helps out a great deal however I am still a bit confused. Anyone have any suggestions on where to get more help with these types of problems? Thanks in advance.

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

14 Dec 2010, 01:30

Expert's post

spyguy wrote:

Great thanks guys. Helps out a great deal however I am still a bit confused. Anyone have any suggestions on where to get more help with these types of problems? Thanks in advance.

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

19 Dec 2010, 20:12

It was a pleasant surprise to see my one year old post still being discussed. Currently I am studying at Stanford GSB. If anyone has any question about Stanford GSB, you may contact me chowdhury_sandip@gsb.stanford.edu

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

31 Dec 2010, 16:56

9.If n is a positive integer and r is the remainder when (n - 1)(n + 1) is divided by 24, what is the value of r ?

(1) n is not divisible by 2.

(2) n is not divisible by 3.

ST 1- if n is not divisible by 2, then n is odd, so both (n - 1) and (n + 1) are even. moreover, since every other even number is a multiple of 4, one of those two factors is a multiple of 4. so the product (n - 1)(n + 1) contains one multiple of 2 and one multiple of 4, so it contains at least 2 x 2 x 2 = three 2's in its prime factorization. But this is not sufficient, because it can be (n-1)*(n+1) can be 2*4 where remainder is 8. it can be 4*6 where the remainder is 0.

ST 2- if n is not divisible by 3, then exactly one of (n - 1) and (n + 1) is divisible by 3, because every third integer is divisible by 3. therefore, the product (n - 1)(n + 1) contains a 3 in its prime factorization. Just like st 1 this is not sufficient

the overall prime factorization of (n - 1)(n + 1) contains three 2's and a 3. therefore, it is a multiple of 24. sufficient

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

01 Jan 2011, 03:21

Expert's post

girishkakkar wrote:

9.If n is a positive integer and r is the remainder when (n - 1)(n + 1) is divided by 24, what is the value of r ?

(1) n is not divisible by 2.

(2) n is not divisible by 3.

ST 1- if n is not divisible by 2, then n is odd, so both (n - 1) and (n + 1) are even. moreover, since every other even number is a multiple of 4, one of those two factors is a multiple of 4. so the product (n - 1)(n + 1) contains one multiple of 2 and one multiple of 4, so it contains at least 2 x 2 x 2 = three 2's in its prime factorization. But this is not sufficient, because it can be (n-1)*(n+1) can be 2*4 where remainder is 8. it can be 4*6 where the remainder is 0.

ST 2- if n is not divisible by 3, then exactly one of (n - 1) and (n + 1) is divisible by 3, because every third integer is divisible by 3. therefore, the product (n - 1)(n + 1) contains a 3 in its prime factorization. Just like st 1 this is not sufficient

the overall prime factorization of (n - 1)(n + 1) contains three 2's and a 3. therefore, it is a multiple of 24. sufficient

Answer C

What if N=1? 1 is not divisible by 2 or 3.

Please discuss specific questions in PS or DS subforums.

As for your question: if n=1 then (n - 1)(n + 1) equals to zero and zero is divisible by every integer (except zero itself) so by 24 too. _________________

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

10 Apr 2011, 11:46

8.If n is a positive integer and r is the remainder when 4 + 7n is divided by 3, what is the value of r ?

(1) n + 1 is divisible by 3.

(2) n > 20

Answer A st1... n+1 divisible by 3..so n=2,5,8,11...... this gives 4+7n=18,39,60....remainder 0 in each case...... st2) insufficient ....n can have any value

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

02 Jul 2011, 06:33

awesome set of questions...VERY VERY helpful _________________

It matters not how strait the gate, How charged with punishments the scroll, I am the master of my fate : I am the captain of my soul. ~ William Ernest Henley

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

07 Aug 2011, 19:25

1

This post received KUDOS

Galiya wrote:

Quote:

st1. take multiples of 8....divide them by 4...remainder =1 in each case...

but how can it be so? multiples of 8 are also multiples of 4

hey

because P has a remainder of 5 after dividing by 8 , so P= (multiple of 8) + 5 take numbers ( 40 + 5 ) / 8 = 5 + ( 5/8 ) divide 45 by 4 the remainder is 1

_________________

How can i lose my faith in life's fairness when i know that the dreams of those who sleep on the feathers are not more beautiful than the dreams of those who sleep on the ground? - Jubran Khaleel Jubran

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

07 Aug 2011, 22:15

Galiya wrote:

HI,Silver!) yep, I missed it. thanks a lot? +1

you are welcome _________________

How can i lose my faith in life's fairness when i know that the dreams of those who sleep on the feathers are not more beautiful than the dreams of those who sleep on the ground? - Jubran Khaleel Jubran

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

07 May 2012, 16:28

[quote="sandipchowdhury"]I have collected these problems on remainder. This type of problem is frequently asked in DS. Answers are also given.

3.If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5.

(2) p is the sum of the squares of two positive integers.

st1. take multiples of 8....divide them by 4...remainder =1 in each case...

st2. p is odd ,since p is square of 2 integers...one will be even and other odd....now when we divide any even square by 4 v ll gt 0 remainder..and when divide odd square vll get 1 as remainder......so intoatal remainder=1 Ans : D

i don t aggree in your approach it is clearly said that P is not a multiple of 8 Since it has a remainder of 5 So your ST 1 ?

FOR sT2 Why do you consider that the sum of 2 square one is even and one odd do you considered that it is the sum of the Following square I am very confused

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

17 Feb 2014, 19:35

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: Collection of remainder problems in GMAT [#permalink]

Show Tags

30 May 2015, 18:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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...