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 \(2^{98} = 256L + N\), where \(L\) and \(N\) are integers and \(0 \le N \le 4\) , what is the value of \(N\) ?

A. 0 B. 1 C. 2 D. 3 E. 4

Given: \(2^{98}=2^8*L+N\) --> divide both parts by \(2^8\) --> \(2^{90}=L+\frac{N}{2^8}\). Now, as both \(2^{90}\) and L are an integers then \(\frac{N}{2^8}\) must also be an integer, which is only possible for \(N=0\) (since \(0\leq{N}\leq4\)).

Re: If 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

08 May 2012, 01:30

Bunuel wrote:

monir6000 wrote:

If \(2^{98} = 256L + N\), where \(L\) and \(N\) are integers and \(0 \le N \le 4\) , what is the value of \(N\) ?

A. 0 B. 1 C. 2 D. 3 E. 4

Given: \(2^{98}=2^8*L+N\) --> divide both parts by \(2^8\) --> \(2^{90}=L+\frac{N}{2^8}\). Now, as both \(2^{90}\) and L are an integers then \(\frac{N}{2^8}\) must also be an integer, which is only possible for \(N=0\) (since \(0\leq{N}\leq4\)).

Answer: A.

Dear Bunuel,

could you pls explain why N/2^8 = 0 ? why it could not be another multiple of 2? L is also there? why we are sure L+N/2^8 do not = 2^90?

If \(2^{98} = 256L + N\), where \(L\) and \(N\) are integers and \(0 \le N \le 4\) , what is the value of \(N\) ?

A. 0 B. 1 C. 2 D. 3 E. 4

Given: \(2^{98}=2^8*L+N\) --> divide both parts by \(2^8\) --> \(2^{90}=L+\frac{N}{2^8}\). Now, as both \(2^{90}\) and L are an integers then \(\frac{N}{2^8}\) must also be an integer, which is only possible for \(N=0\) (since \(0\leq{N}\leq4\)).

Answer: A.

Dear Bunuel,

could you pls explain why N/2^8 = 0 ? why it could not be another multiple of 2? L is also there? why we are sure L+N/2^8 do not = 2^90?

Sure. Given that \(0\leq{N}\leq4\), so N can only be 0, 1, 2, 3, or 4. So, N/2^8 is an integer only for N=0.

Re: If 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

20 Dec 2012, 03:02

Another approach to tackle this problem: The given relation is: \(2^98=256L+N\) This is actually written in "remainder format" i.e. X=Quotient*some integer + Remainder. Here X=\(2^98\), Quotient=256 and Remainder=0.

We just have to check whether \(256\) OR \(2^8\) divides \(2^ 98\). If yes, then the value of N is straightaway 0. No need to check choices even. _________________

Re: If 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

27 May 2013, 16:02

Marcab wrote:

Another approach to tackle this problem: The given relation is: \(2^98=256L+N\) This is actually written in "remainder format" i.e. X=Quotient*some integer + Remainder. Here X=\(2^98\), Quotient=256 and Remainder=0.

We just have to check whether \(256\) OR \(2^8\) divides \(2^ 98\). If yes, then the value of N is straightaway 0. No need to check choices even.

Could you tell me if I'm understanding it correctly?

What if we had 12/5 = 2 remainder 2? Then in remainder format we would have 2 * 5 + 2 = 12.

So according to what you're saying, if we just have to check if 2 can divide 12? In this case it does, but the remainder is 2 not 0.

Please, help me in understanding where I'm going wrong.

Re: If 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

28 May 2013, 01:36

youngkacha wrote:

Marcab wrote:

Another approach to tackle this problem: The given relation is: \(2^98=256L+N\) This is actually written in "remainder format" i.e. X=Quotient*some integer + Remainder. Here X=\(2^98\), Quotient=256 and Remainder=0.

We just have to check whether \(256\) OR \(2^8\) divides \(2^ 98\). If yes, then the value of N is straightaway 0. No need to check choices even.

Could you tell me if I'm understanding it correctly?

What if we had 12/5 = 2 remainder 2? Then in remainder format we would have 2 * 5 + 2 = 12.

So according to what you're saying, if we just have to check if 2 can divide 12? In this case it does, but the remainder is 2 not 0.

Please, help me in understanding where I'm going wrong.

Your reasoning unfortunately is incorrect since 5 is the divisor in question. The first 2 obtained is the quotient (which is the integer result obtained when dividing the numerator with the denominator) and cannot be used to question whether 2 divides 12 or not.

To further illustrate, take a look at the problem itself : \(2^98 = 256*L + N\) which can be rephased as the following : does 256 divide \(2^98\) ? (Since in this instance \(2^98\) is the numerator, 256 is the denominator, L is the quotient and N is the remainder). To which the answer, obviously, is yes, since 256 can be written as a power of two, in this case 2^8.

Re: If 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

28 May 2013, 16:44

Virgilius wrote:

youngkacha wrote:

Marcab wrote:

Another approach to tackle this problem: The given relation is: \(2^98=256L+N\) This is actually written in "remainder format" i.e. X=Quotient*some integer + Remainder. Here X=\(2^98\), Quotient=256 and Remainder=0.

We just have to check whether \(256\) OR \(2^8\) divides \(2^ 98\). If yes, then the value of N is straightaway 0. No need to check choices even.

Could you tell me if I'm understanding it correctly?

What if we had 12/5 = 2 remainder 2? Then in remainder format we would have 2 * 5 + 2 = 12.

So according to what you're saying, if we just have to check if 2 can divide 12? In this case it does, but the remainder is 2 not 0.

Please, help me in understanding where I'm going wrong.

Your reasoning unfortunately is incorrect since 5 is the divisor in question. The first 2 obtained is the quotient (which is the integer result obtained when dividing the numerator with the denominator) and cannot be used to question whether 2 divides 12 or not.

To further illustrate, take a look at the problem itself : \(2^98 = 256*L + N\) which can be rephased as the following : does 256 divide \(2^98\) ? (Since in this instance \(2^98\) is the numerator, 256 is the denominator, L is the quotient and N is the remainder). To which the answer, obviously, is yes, since 256 can be written as a power of two, in this case 2^8.

Sorry, but I'm still lost. Marcab has the quotient as 256, but you're saying 256 is the divisor and L is the quotient?

Re: If 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

19 Jul 2014, 02:36

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 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

26 Nov 2015, 06:57

Bunuel wrote:

monir6000 wrote:

If \(2^{98} = 256L + N\), where \(L\) and \(N\) are integers and \(0 \le N \le 4\) , what is the value of \(N\) ?

A. 0 B. 1 C. 2 D. 3 E. 4

Given: \(2^{98}=2^8*L+N\) --> divide both parts by \(2^8\) --> \(2^{90}=L+\frac{N}{2^8}\). Now, as both \(2^{90}\) and L are an integers then \(\frac{N}{2^8}\) must also be an integer, which is only possible for \(N=0\) (since \(0\leq{N}\leq4\)).

Answer: A.

I'm sorry but if you're dividing both sides of the equation by \(2^8\) won't it cancel on the right side of the equation? So how do we come to \(\frac{N}{2^8}\) ? _________________

If \(2^{98} = 256L + N\), where \(L\) and \(N\) are integers and \(0 \le N \le 4\) , what is the value of \(N\) ?

A. 0 B. 1 C. 2 D. 3 E. 4

Given: \(2^{98}=2^8*L+N\) --> divide both parts by \(2^8\) --> \(2^{90}=L+\frac{N}{2^8}\). Now, as both \(2^{90}\) and L are an integers then \(\frac{N}{2^8}\) must also be an integer, which is only possible for \(N=0\) (since \(0\leq{N}\leq4\)).

Answer: A.

I'm sorry but if you're dividing both sides of the equation by \(2^8\) won't it cancel on the right side of the equation? So how do we come to \(\frac{N}{2^8}\) ?

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...