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: Remainder when k^2/8? [#permalink]
19 Jun 2010, 19:12

My initial thoughts were: 1) \(k = 2n -1\), so \(k\) must be odd 2) For \(k^2\) to be divisible by 8, \(k^2\) must contain at least 3 2's. Therefore, each k must contain 2 2's.

I'm not sure how to continue using my initial thoughts.

The official answer shows: i) Express \(k^2 = (2n-1)^2 = 4n^2 - 4n + 1\)

ii) Factor to \(k = 4n(n-1)+1\)

Step ii doesn't make sense to me. If you factor out \(4n\) on the right side, why would the left not still be \(k^2\)? If this is just a misprint, then the the next step makes sense to me.

iii) If \(n\) is even, then \(n-1\) is odd, while if \(n\) is odd, then \(n-1\) is even. Therefore no matter what integer \(n\) is, \(k\) will equal 4 * even * odd, plus 1. In other words, \(k\) will equal a multiple of 8, plus 1. Therefore, the remainder of \(k^2/8\) is 1.

Re: Remainder when k^2/8? [#permalink]
20 Jun 2010, 07:18

3

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

jpr200012 wrote:

If \(k = 2n - 1\), where \(n\) is an integer, what is the remainder of \(k^2/8\)?

A. 1 B. 3 C. 5 D. 7 E. Cannot be determined from the information given.

This one can be done very easily with number picking. As you correctly noted \(k = 2n - 1\) means that k is an odd number (basically \(k = 2n - 1\) is a formula of an odd number).

Now let's try several odd numbers: k=1 --> k^2=1 ---> remainder upon division of 1 by 8 is 1; k=3 --> k^2=3 ---> remainder upon division of 9 by 8 is 1; k=5 --> k^2=25 ---> remainder upon division of 25 by 8 is 1;

At this point we can safely assume that this will continue for all odd numbers.

But if you want algebraic approach, here you go: \(k = 2n - 1\) --> \(k^2=(2n-1)^2=4n^2-4n+1=4n(n-1)+1\) --> what is a remainder when \(4n(n-1)+1\) is divided by 8:

Now either \(n\) or \(n-1\) will be even so in any case \(4n(n-1)=4*odd*even=multiple \ of \ 8\), so \(4n(n-1)\) is divisible by 8, so \(4n(n-1)+1\) divided by 8 gives remainder of 1.

Re: Remainder when k^2/8? [#permalink]
20 Jun 2010, 21:10

Hi Bunuel,

How come 1 div by 8 gives remainder as 1???

utin.

Bunuel wrote:

jpr200012 wrote:

If \(k = 2n - 1\), where \(n\) is an integer, what is the remainder of \(k^2/8\)?

A. 1 B. 3 C. 5 D. 7 E. Cannot be determined from the information given.

This one can be done very easily with number picking. As you correctly noted \(k = 2n - 1\) means that k is an odd number (basically \(k = 2n - 1\) is a formula of an odd number).

Now let's try several odd numbers: k=1 --> k^2=1 ---> remainder upon division of 1 by 8 is 1; k=3 --> k^2=3 ---> remainder upon division of 9 by 8 is 1; k=5 --> k^2=25 ---> remainder upon division of 25 by 8 is 1;

At this point we can safely assume that this will continue for all odd numbers.

But if you want algebraic approach, here you go: \(k = 2n - 1\) --> \(k^2=(2n-1)^2=4n^2-4n+1=4n(n-1)+1\) --> what is a remainder when \(4n(n-1)+1\) is divided by 8:

Now either \(n\) or \(n-1\) will be even so in any case \(4n(n-1)=4*odd*even=multiple \ of \ 8\), so \(4n(n-1)\) is divisible by 8, so \(4n(n-1)+1\) divided by 8 gives remainder of 1.

Re: Remainder when k^2/8? [#permalink]
21 Jun 2010, 01:52

1

This post received KUDOS

Expert's post

utin wrote:

Hi Bunuel,

How come 1 div by 8 gives remainder as 1???

utin.

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 (8 in our case) is more than dividend (1 in our case) then the reminder equals to the dividend:

1 divided by 8 yields a reminder of 1 --> \(1=0*8+1\); or:

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

Re: Remainder when k^2/8? [#permalink]
09 Aug 2013, 05:36

1

This post received KUDOS

K=2n-1

K^2 = 4n^2 +1 - 4n

Dividing both sides by '8'.

=> K^2/8 = 4(n^2-n)/8 +1/8

=> n(n-1)/2 + 1/8

In both cases whether 'n' is EVEN or ODD n(n-1) is divisible by '2'.

Hence,

Remainder is = 1/8 = 1 _________________

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

Re: If k = 2n - 1, where n is an integer, what is the remainder [#permalink]
03 Sep 2014, 05:18

1

This post received KUDOS

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. _________________

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...