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:

\(2^{1927}=(2^5)^{385}\cdot2^2=(33-1)^{385}\cdot{4}=(M33-1)\cdot{4}=M33-4=M33-33+29=M33+29\). (\(M33\) designates multiple of \(33\).)

\(2^{1929}=2^{1927}\cdot2^2=(M33+29)\cdot{4}=M132+29\cdot{4}=M132+116\). \(M33+29\) means \(33n+29\), for some integer \(n\). Then \((M33+29)\cdot{4}=132n+29\cdot{4}=M132+116\).

Hence, the remainder is 116.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

\(2^{1929}=2^{1927}\cdot2^2=(M33+29)\cdot{4}=M132+29\cdot{4}=M132+116\). \(M33+29\) means \(33n+29\), for some integer \(n\). Then \((M33+29)\cdot{4}=132n+29\cdot{4}=M132+116\).

Hence, the remainder is 116.

What is wrong if i cancel out 2^2 in the numerator and divide by 33 . I still have to get the same reminder right?

\(2^{1927}=(2^5)^{385}\cdot2^2=(33-1)^{385}\cdot{4}=(M33-1)\cdot{4}=M33-4=M33-33+29=M33+29\). (\(M33\) designates multiple of \(33\).)

\(2^{1929}=2^{1927}\cdot2^2=(M33+29)\cdot{4}=M132+29\cdot{4}=M132+116\). \(M33+29\) means \(33n+29\), for some integer \(n\). Then \((M33+29)\cdot{4}=132n+29\cdot{4}=M132+116\).

\(2^{1929}=2^{1927}\cdot2^2=(M33+29)\cdot{4}=M132+29\cdot{4}=M132+116\). \(M33+29\) means \(33n+29\), for some integer \(n\). Then \((M33+29)\cdot{4}=132n+29\cdot{4}=M132+116\).

Hence, the remainder is 116.

What is wrong if i cancel out 2^2 in the numerator and divide by 33 . I still have to get the same reminder right?

Hi g3kr,

Both of the last 2 solutions fetching 29 and 116 are correct processwise, but for a logical mistake in former.

Consider the case, 16/10 will have a remainder of 6, but its not same as 8/5 which has a remainder of 3 only.

The difference? notice that for any y/x remainder can not be greater than x, hence if you have factored out a number from numerator and denominator, you have reduced remainder by that times. In our example of 16/10: if we make it 8/5 by dividing each term by 2 and get 3 as remainder. we should multiply result by 2 to get correct ans 6.

hence, once you arrived in first solution (M33+29)

you should remember to multiply it by 4, the one which was factored out earlier. Or better dont factor out and cancel a term in a remainder problem.

What is wrong if i cancel out 2^2 in the numerator and divide by 33 . I still have to get the same reminder right?[/quote]

Hi g3kr,

Both of the last 2 solutions fetching 29 and 116 are correct processwise, but for a logical mistake in former.

Consider the case, 16/10 will have a remainder of 6, but its not same as 8/5 which has a remainder of 3 only.

The difference? notice that for any y/x remainder can not be greater than x, hence if you have factored out a number from numerator and denominator, you have reduced remainder by that times. In our example of 16/10: if we make it 8/5 by dividing each term by 2 and get 3 as remainder. we should multiply result by 2 to get correct ans 6.

hence, once you arrived in first solution (M33+29)

you should remember to multiply it by 4, the one which was factored out earlier. Or better dont factor out and cancel a term in a remainder problem.

Hope it helps.[/quote]

Thanks,

Even I had the same ambiguity.. But just to be sure.. there wont be both options in the answer : 29 & 116 ?

Even I had the same ambiguity.. But just to be sure.. there wont be both options in the answer : 29 & 116 ?

Both 29 and 116 can be in the options because the answer is only 116. In fact, it is quite likely that both will be there since if there is a way to trip you, GMAT will not let it go. You can choose to cancel off terms in the numerator and denominator or you can choose to keep them as it is. In either case, answer will be 116.

\(\frac{8^{643}}{132} = \frac{8*8^{642}}{132}\)

We get \(2*\frac{8^{642}}{33} = 2*\frac{2^{1926}}{33}\)

What is wrong if i cancel out 2^2 in the numerator and divide by 33 . I still have to get the same reminder right?

Hi g3kr,

Both of the last 2 solutions fetching 29 and 116 are correct processwise, but for a logical mistake in former.

Consider the case, 16/10 will have a remainder of 6, but its not same as 8/5 which has a remainder of 3 only.

The difference? notice that for any y/x remainder can not be greater than x, hence if you have factored out a number from numerator and denominator, you have reduced remainder by that times. In our example of 16/10: if we make it 8/5 by dividing each term by 2 and get 3 as remainder. we should multiply result by 2 to get correct ans 6.

hence, once you arrived in first solution (M33+29)

you should remember to multiply it by 4, the one which was factored out earlier. Or better dont factor out and cancel a term in a remainder problem.

Hope it helps.

Thanks,

Even I had the same ambiguity.. But just to be sure.. there wont be both options in the answer : 29 & 116 ?

Hi mindmind (too many minds :D),

Exactly what I explained with example of 16/10 and 8/5. if i ask you remainder of 16/10 and give u option of 6 and 3, will you be confused? I dont think so

If u factor out a number, the "fraction" remains same but "remainder" does not. please re-read the post. Also this is why it is important not to factor out in remainder problem. Also try to check one more way of understanding it: a remainder is shown as below: if n = qm +r then kn = kqm +kr where Kn is a multiple of n and km is multiple of m. But this changes remainder from r to kr.

Re: What is the remainder when 8^643/132? [#permalink]

Show Tags

12 Feb 2014, 03:29

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: What is the remainder when 8^643/132? [#permalink]

Show Tags

13 Feb 2016, 05:49

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.
_________________

\(2^{1927}=(2^5)^{385}\cdot2^2=(33-1)^{385}\cdot{4}=(M33-1)\cdot{4}=M33-4=M33-33+29=M33+29\). (\(M33\) designates multiple of \(33\).)

\(2^{1929}=2^{1927}\cdot2^2=(M33+29)\cdot{4}=M132+29\cdot{4}=M132+116\). \(M33+29\) means \(33n+29\), for some integer \(n\). Then \((M33+29)\cdot{4}=132n+29\cdot{4}=M132+116\).

Hence, the remainder is 116.

I got the correct answer but took me very long to do it, which would be tricky in exam conditions. is this really a gmat question??

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...