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 remainder, when (n-1)(n+1) is divided by 24 what is value of r?

1) n is not divisible by 2 2) n is not divisible by 3

C

(1) n is odd if n is odd, then pick some numbers... (n, (n-1)*(n+1), r) (1, 0, 0) (3, 8, 8) INSUFFICIENT

(2) (1, 0, 0) (2, 3, 3) INSUFFICIENT

Together, find some more examples... (1, 0, 0) (3, 8, 8) (5, 24, 0) (7, 48, 0) (9, 56, 8) So the pattern is if n is divisible by 3, then remainder is 8, if not remainder is 0. Together is SUFFICIENT

(3, 8, 8) So the pattern is if n is divisible by 3, then remainder is 8, if not remainder is 0. Together is SUFFICIENT

Do you misconceive the circumstance? Cos' (n-1)(n+1) is divided by 24, the remainder is r, thus (n-1)(n+1)>24. Further, should we try many numbers to answer whether or not the given information is sufficient? I think we'll conclude only that one case is not sufficient if we find a number counts against the case, but we can not conclude that one case is sufficient if we find a number does not count against the case. That why I cast doubt on C.

(3, 8, 8) So the pattern is if n is divisible by 3, then remainder is 8, if not remainder is 0. Together is SUFFICIENT

Do you misconceive the circumstance? Cos' (n-1)(n+1) is divided by 24, the remainder is r, thus (n-1)(n+1)>24. Further, should we try many numbers to answer whether or not the given information is sufficient? I think we'll conclude only that one case is not sufficient if we find a number counts against the case, but we can not conclude that one case is sufficient if we find a number does not count against the case. That why I cast doubt on C.

Need a general solution!

(n-1)*(n+1) can be less than 24. For example, fraction 1/4 has a remainder of 1. Regarding how many numbers we should try, if you are able to figure out the pattern, that should be sufficient.

N is not divisible by 2: Means n-1 and n+1 are even. And - As they are consecutive numbers one of them is divisible by 2 and other by 4. So: (n-1)* (N+1) is divisible by 8 - not sufficient to conclude that the product is divisible by 24

N is not divisible by 3: means as n-1,n,n+1 are consecutive , one of them: (n-1) or (n+1) is divisible by 3. Again, not sufficient for 24.

Together - (n-1)(n+1) is divisible by 2,4,3 i.e: 2*4*3 = 24 . Hence, r=0

N is not divisible by 2: Means n-1 and n+1 are even. And - As they are consecutive numbers one of them is divisible by 2 and other by 4. So: (n-1)* (N+1) is divisible by 8 - not sufficient to conclude that the product is divisible by 24

N is not divisible by 3: means as n-1,n,n+1 are consecutive , one of them: (n-1) or (n+1) is divisible by 3. Again, not sufficient for 24.

Together - (n-1)(n+1) is divisible by 2,4,3 i.e: 2*4*3 = 24 . Hence, r=0

does it make sense?

I think this very much makes sense, thanks.

I got this question on GMATPrep (it's roughly a Q50 question, give or take a few), and was able to solve it by just plugging in numbers, but it's very helpful to know where the logic comes from - much easier and quicker to solve!

This is how I solved it From statement 1) 2 is not a factor of n ( we can conclude that n is odd ) so by plugging in numbers like n=15 remainder=1 and for n=17 remainder =0 i.e not sufficient From statement 2 it says that 3 is not a factor of n so we can have n as an even number or it can be prime number( not #0 again by plugging in values we can find that the remainders are different.i.e insufficient.

So by combining both we can conclude that the value for n will be a prime number which is not 2 and 3 for n=19 remainder =0 ; for n=13 remainder =0 ( just testing cases )

This was my approach. Did you actually calculate out the remainders when you tried n=15 and 17? I just used n=2 or n=5 to arrive at my examples. C gives you that n has to be prime other than 2 and 3 and you test a few cases to find that all non 2 and 3 primes will provide a 0 remainder for N.

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

1) n is not divisible by 2 2) n is not divisible 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?

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: If n is a positive integer and r is remainder, when [#permalink]

Show Tags

14 May 2015, 01:57

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 remainder, when [#permalink]

Show Tags

17 May 2015, 21:51

rite2deepti wrote:

This is how I solved it From statement 1) 2 is not a factor of n ( we can conclude that n is odd ) so by plugging in numbers like n=15 remainder=1 and for n=17 remainder =0 i.e not sufficient From statement 2 it says that 3 is not a factor of n so we can have n as an even number or it can be prime number( not #0 again by plugging in values we can find that the remainders are different.i.e insufficient.

So by combining both we can conclude that the value for n will be a prime number which is not 2 and 3 for n=19 remainder =0 ; for n=13 remainder =0 ( just testing cases )

SO answer is C

Guys let me know if my solution is good ?

Great....That,s how I solved this question.... _________________

Freedom is not a gift...It is a responsibility to pass on.....

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...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...

By Libby Koerbel Engaging a room of more than 100 people for two straight hours is no easy task, but the Women’s Business Association (WBA), Professor Victoria Medvec...