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:

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.

Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS [#permalink]

Show Tags

03 Jul 2010, 22:54

Question 2:

To be divisible by 24, a number must be divisible by 8 and by 3.

Statement 1: n is not divisible by 2. This basically says that n is an odd number. So if n is an odd number, both n-1 and n+1 are even numbers, which means that (n-1)(n+1) must be divisible by 4.

n-1 = 2k n+1 = 2p

Between any two consecutive even integers, one has to be divisible by 4, so we know that the given number is divisible by 8.

Statement 2: n is not divisible by 3. However, between any three consecutive numbers one has to be divisible by 3. Considering the consecutive numbers (n-1)(n)(n+1) we know that n isn't divisible by 3, so either (n-1) or (n+1) should be divisible by 3.

Combining both together, we know that (n-1)(n+1) is divisible by 8 and 3 and hence 24. So reminder = 0. Hence answer is C.

Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)

Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS [#permalink]

Show Tags

04 Jul 2010, 05:21

in question 2, statement 2 alone will not be sufficient unless taken in conjunction with stmt 1 because if we consider ONLY stmt 2 - then 10 is a value which is NOT divisible by 3 and (10-1)*(10+1) = gives 99 which does not leave a remainder 0 when divided by 24. Thus IMO the answer to problem 2 should be A - stmt 1 alone is sufficient... pls correct me if I am wrong.

Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS [#permalink]

Show Tags

06 Jul 2010, 23:58

whiplash2411 wrote:

Question 2:

To be divisible by 24, a number must be divisible by 8 and by 3.

Statement 1: n is not divisible by 2. This basically says that n is an odd number. So if n is an odd number, both n-1 and n+1 are even numbers, which means that (n-1)(n+1) must be divisible by 4.

n-1 = 2k n+1 = 2p

Between any two consecutive even integers, one has to be divisible by 4, so we know that the given number is divisible by 8.

Statement 2: n is not divisible by 3. However, between any three consecutive numbers one has to be divisible by 3. Considering the consecutive numbers (n-1)(n)(n+1) we know that n isn't divisible by 3, so either (n-1) or (n+1) should be divisible by 3.

Combining both together, we know that (n-1)(n+1) is divisible by 8 and 3 and hence 24. So reminder = 0. Hence answer is C.

n is not divisble by 2 implies n-1 and n+1 are even , this implies the product of these two numbers i divisible by 4 and not 8. both the statements taken together say that the number is not even and not divisible by 3 , meaning all odd numbers not divisible by 3 . these are 1,5,7,11,13,17--- , if u (square of any of these minus 1 )/24 = 1 and not 0

@bunuel... 3 divided by 24 yields remainder of 3???

ruchichitral wrote:

bunnel:- i am too confused....how come 3 divided by 24 the remainder is 3?

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

So when divisor (24 in our case) is more than dividend (3 in our case) then the reminder equals to the dividend:

3 divided by 24 yields a reminder of 3 --> \(3=0*24+3\);

or:

5 divided by 6 yields a reminder of 5 --> \(5=0*6+5\).

Re: If n is a positive integer and r is the remainder when (n-1) [#permalink]

Show Tags

01 Dec 2014, 14:09

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: If n is a positive integer and r is the remainder when (n-1) [#permalink]

Show Tags

03 Oct 2017, 10:20

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.
_________________