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: Data sufficiency remainder problem [#permalink]

Show Tags

26 Apr 2010, 13:56

1

This post received KUDOS

ksharma12 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.

I know this is not new to the forum, but I couldnt find an explanation that was clear to me. Can someone provide detailed reasoning?

Thank you

For statement 1. take n = 5 the number becomes 4*6 when divided by 24 gives remainder 0 take n = 7 the number becomes 6*8 when divided by 24 gives remainder 0 take n = 9 the number becomes 8*10 when divided by 24 gives remainder different than 0. Thus not sufficient.

For statement 2. take n= 5 and n=7 it will give remainder as 0 if you take n = 12 14 number becomes 13*15 when divided by 24 gives remainder other than 0. Thus not sufficient.

So only answers left are either C or E

Now you must have observed that when n is not multiple of 2 or not multiple of 3 independently we are unable to answer the question. But when n is neither multiple of 2 nor 3 then we are able to answer the question as r=0 in all those cases.

Hence C

Another way of looking at it is.... take n-1 , n , n+1 these are 3 consecutive integers, we know we will have both the multiple of 2 and 3 in three consecutive integer. If that integer is same i.e. n is multiple of both 2 and 3 , then n-1 and n+1 will never have any factor for 24.

But if n is not divisible by both 2 and 3 that means n-1 and n+1 will be both be even with one of them having 3 as factor. One another point to be noted is , since n-1 and n+1 are consecutive even numbers that implies one of them must be divisible by 4.

so (n-1)(n+1) will be always divisible by 2*4*3 = 24 hence r will always remain as 0 (When n is neither divisible by 2 nor by 3)
_________________

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?

Plug-in 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.

Re: If n is a positive integer and r is the remainder when (n - [#permalink]

Show Tags

26 Sep 2012, 12:44

ksharma12 wrote:

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 must be odd. Try n = 1 and n = 3. Not sufficient.

(2) Try n = 1 and n = 2. Not sufficient.

(1) and (2) together: When divided by 6, the remainders can be 0, 1, 2, 3, 4, 5. Since n is not divisible neither by 2, nor by 3, when divide by 6, n can give remainder 1 or 5, which means n is of the form 6k + 1 or 6k - 1, for some positive integer k. If n = 6k + 1, then (n - 1)(n + 1) = 6k(6k + 2) = 12k(3k + 1) = 12k(k + 2k + 1). Since 2k + 1 is odd, k and 3k + 1 are of different parities, so their product is for sure divisible by 2. Altogether, (n - 1)(n + 1) is divisible by 24. If n = 6k - 1, then (n - 1)(n + 1) = (6k - 2)6k)= 12k(3k - 1) = 12k(k + 2k - 1). Since 2k - 1 is odd, k and 3k - 1 are of different parities, so their product is for sure divisible by 2. Altogether, (n - 1)(n + 1) is divisible by 24. The remainder must be 0. Sufficient.

Answer C.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If n is a positive integer and r is the remainder when (n - [#permalink]

Show Tags

03 Feb 2016, 10:47

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.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...