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: PS: Exponents (m06q07) [#permalink]
29 Mar 2010, 21:11

5

This post received KUDOS

\(2^{98} = 256L + N\) The given expression is a classic example of division: \(dividend (2^{98}) = quotient (L)*divisor (256=2^8) + remainder (N)\). Since \(2^{98}/2^8=2^{90}=L\) the remainder \(N=0\).

Re: PS: Exponents (m06q07) [#permalink]
30 Mar 2010, 00:28

I saw it today ( my country time is CET ) the number inside brackets 1 not represent N

The last digit of 2*98 is 4 ( I have sent a theory about last digit few months ago ) This happen a) when 256∙L have last digit 4 and N=0 ( ..6 ∙ ..4 give 4, this happen L=2*90,=2*2) b) when 256∙L have last digit 2 and N=2 (...6 ∙ ..2 give 2, this happen L=2*89 =2*1)

I tried to compare it with a similar example of less power like 2^6 = 2^3L + N and reached the conclusion that it can only be 0. though I liked nvgroshar's reasoning better !

In options 2 and 4, the number is a FRACTION, and not a odd/even division,

Like L = 2^98 - 2 / 2^8 will run from 2^n/2^m exponencial divison so will be a fraction. Example = 2^8 - n / 2^4 to n = 0 , 2^8-0/2^4=256/16=16.000 to n = 2 , 2^8-2/2^4=256-2/16=254/16=15.875

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 \(0\leq{N}\leq4\)).

I was able to deduce the part where 256 = 2^90 and then get the 2^8 but I did not realize that this equation was the "dividend=quotient * divisor + remainder." This was a tricky question for sure and I guess now after seeing the solution I understand why A=0. I'll have to put this in my error log so I know why in the future

\(2^{98} = 256L + N\) The given expression is a classic example of division: \(dividend (2^{98}) = quotient (L)*divisor (256=2^8) + remainder (N)\). Since \(2^{98}/2^8=2^{90}=L\) the remainder \(N=0\).

Lovely way to arrive at the answer _________________