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 350,000 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: Data Sufficiency with remainder [#permalink]
04 Aug 2010, 02:54

3

This post received KUDOS

Expert's post

kwhitejr wrote:

Can anyone demonstrate the following?

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) 2 is not a factor of n (2) 3 is not a factor of n

Hi, and welcome to Gmat Club! Below is the solution for your problem:

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?

Number plugging method:

(n-1)(n+1)=n^2-1

(1) n is not divisible by 2 --> pick two odd numbers: let's say 1 and 3 --> if n=1, then n^2-1=0 and as zero is divisible by 24 (zero is divisible by any integer except zero itself) so remainder is 0 but if n=3, then n^2-1=8 and 8 divided by 24 yields remainder of 8. Two different answers, hence not sufficient.

(2) n is not divisible by 3 --> pick two numbers which are not divisible by 3: let's say 1 and 2 --> if n=1, then n^2-1=0, so remainder is 0 but if n=2, then n^2-1=3 and 3 divided by 24 yields remainder of 3. Two different answers, hence not sufficient.

(1)+(2) Let's check for several numbers which are not divisible by 2 or 3: n=1 --> n^2-1=0 --> remainder 0; n=5 --> n^2-1=24 --> remainder 0; n=7 --> n^2-1=48 --> remainder 0; n=11 --> n^2-1=120 --> remainder 0. Well it seems that all appropriate numbers will give remainder of 0. Sufficient.

Algebraic approach:

(1) n is not divisible by 2. Insufficient on its own, but this statement says that n=odd --> n-1 and n+1 are consecutive even integers --> (n-1)(n+1) must be divisible by 8 (as both multiples are even and one of them will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8, 10); (10, 12), ...).

(2) n is not divisible by 3. Insufficient on its own, but form this statement either n-1 or n+1 must be divisible by 3 (as n-1, n, and n+1 are consecutive integers, so one of them must be divisible by 3, we are told that it's not n, hence either n-1 or n+1).

(1)+(2) From (1) (n-1)(n+1) is divisible by 8, from (2) it's also divisible by 3, therefore it must be divisible by 8*3=24, which means that remainder upon division (n-1)(n+1) by 24 will be 0. Sufficient.

hi all, please help solving the following question...

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

1) n is not divisible by 2 => n is not even.

n could be 1, 3, 5, 7, 9

and (n+1) * (n-1) could be different values.

2) n is not divisible by 3.

n could be 1, 2, 4, 5, 7, 8, 10...

again (n+1) * (n-1) could be different values.

Combine (1) and (2), n could be 1, 5, 7, 11, 13; ((n+1), (n-1)) would contain a multiple of 6 and 4 when n is 5, 7, 11, 13 ..

When n =1, (n-1) is zero and the remainder is zero.

Multiple of 6 and 4 would contain (3*2) and (2^2) = > (3 * 2 * 2 *2) = > 24.

(n-1)*(n+1) could be 4*6 or 6*8 and either way the remainder when (n-1)*(n+1)/24 would be 0.

Hence both statements is sufficient to answer this. _________________

Support GMAT Club by putting a GMAT Club badge on your blog

PS: This way of solving would probably be better in a GMAT hall with a 2 minute time constraint. A rigorous algebraic proof can also be derived otherwise.

Re: Data Sufficiency with remainder [#permalink]
15 Sep 2010, 16:44

I got this problem right (on the GMATPrep #1 CAT) but I'm sure it took me at least 3 minutes, if not closer to 5. At least a minute to set up the problem and figure out how to approach it (the way described above is what I first thought of, but I tried to think of easier ways), another minute to test numbers, and another minute to think of ways on how my test could be wrong.

Any suggestions on how to cut down on timing for unfamiliar problems like these? These divisor/remainder problems always get to me. _________________

Re: Data Sufficiency with remainder [#permalink]
16 Sep 2010, 23:02

Folks,

Consider this approach.

Stmnt 1: N is not divisible by 2 ==> n=2k+1

==> n-1 * n+1 = 2k(2k+2)=4K(k+1) te K(K+1) is always divisible by 2 as they are consecutive integers. hence 4K(k+1) is always divisible by 8, but may not be by 24 (i.e. 8*3)....NOT Suff. Summary from stmnt 1: (n-1)*(n+1) is always divisible by 8

Stmnt 2: N is not divisible by 3 ==> 2 cases -> remainder can be 1 or 2 ==> n=3k+1 or 3K+2

with n=3K+1 ==> ==> n-1 * n+1 = 3k(3k+2)=is divisible by 3 but not by 2 for ODD K: hence not divisible by 24 (i.e. 2^3 * 3) with n=3K+2 ==> ==> n-1 * n+1 = (3k+1)(3k+3)=3(3k+1)(K+1) is divisible by 3 but not by 2 for EVEN K Not suff. Summary from stmnt 2: (n-1)*(n+1) is always divisible by 3

1 & 2 summaries. (n-1)*(n+1) is always divisible by 8*3 = 24..Suff.

Answer "C"

always divisible by 2 as they are consecutive integers. hence 4K(k+1) is always divisible by 8, but may not be by 24 (i.e. 8*3)....NOT Suff.

Divisiblity remainder [#permalink]
10 Feb 2011, 23:37

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) 2 is not a factor of n. (2) 3 is not a factor of

I marked ans e as we cannot find one value while using two other prime 5 & 7. So defainete value of r can be find out. But the ans is c. can anyone confim

Re: Divisiblity remainder [#permalink]
11 Feb 2011, 00:16

24 has the factors; 2*2*2*3

If (n-1)(n+1) contains at least the above factors, it will be divisible by 24.

1. Means; n is odd; and n-1 and n+1 are consecutive even; thus one of them will have at least 2 as factor and the other at least 2*2 Thus; we know (n-1)(n+1) contains 2*2*2 as factor. But, it may or may not contain 3 as its factor.

Not sufficient.

2. if n is not divisible by 3; one of (n-1)(n+1) must be divisible by 3. One number of any set of 3 consecutive numbers is always divisible by 3. We know n is not divisible and n-1,n,n+1 are consecutive. if n is not divisible by 3, one of (n-1) and (n+1) must be divisible by 3 or have a factor as 3. However, we don't know whether these numbers have 2*2*2 as factors.

Not sufficient.

Combining both; (n-1)(n+1) has factors 2*2*2*3 and must be divisible by 24. Remainder 0.

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: If n is a positive integer and r is the remainder when (n-1) [#permalink]
02 Oct 2014, 08:06

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. _________________

hey guys, A metallurgist but currently working in a NGO and have scheduled my GMAT in December for second round .....u know. I read some but valuable blogs on this...

Today, 1st year Rotman students had a great simulation event hosted by Scotiabank, one of Canada’s best and largest banks. Attended by entire Rotman 1st year students, the...

Nope. I never learned finance ever in my life until I came to Rotman. This is why I got really frustrated when this term started because I was certain...