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 problem. [#permalink]
16 Oct 2012, 20:11

1

This post was BOOKMARKED

Actually the cyclicity is 3:

(2^0)/7 = 0 R 1 (2^1)/7 = 0 R 2 (2^2)/7 = 0 R 4 (2^3)/7 = 1 R 1 (2^4)/7 = 2 R 2 (2^5)/7 = 4 R 4 ...

The pattern is such that the remainder is 4 for every third one and we know there are 201 numbers between 0 and 200. Since 201 is a multiple of 3, D is in fact correct.

The remainder of 4^40 would be same as that of 4 . Is my approach wrong? Can you provide similar examples, where it is not.

In this case, \(4^{40}\) is a \(M7+4\) : \(\,\,4^{40}=2^{80}\) and \(80=M3+2\) (the cycle is 3, because \(2^{3}=8=M7+1\)). \(4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4\). Or - \(4^3=64=M7+1\), therefore \(4^{40}=(4^3)^{13}\cdot{4}=(M7+1)^{13}\cdot{4}=(M7+1)\cdot{4}=M7+4\).

Consider for example \(5^{40}\): \(5=7\cdot{0}+5=M7+5\) and \(5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6\). \(5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2\) and not \(M7+5\). _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 17 Oct 2012, 02:48, edited 1 time in total.

Re: Remainder problem. [#permalink]
17 Oct 2012, 02:43

Please help me understand better :

The remainder of 4^40 would be same as that of 4 . Is my approach wrong? Can you provide similar examples, where it is not.[/quote]

In this case, \(4^{40}\) is a \(M7+4\) : \(\,\,4^{40}=2^{80}\) and \(80=M3+2\) (the cycle is 3, because \(2^{3}=8=M7+1\)). \(4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4\).

Consider for example \(5^{40}\): \(5=7\cdot{0}+5=M7+5\) and \(5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6\). \(5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2\) and not \(M7+5\).[/quote]

Yes, Agreed So I should consider : Something near to Remainder 1

Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime eg : 4^40 divided by 7 give a remainder of 4.

Re: Remainder problem. [#permalink]
17 Oct 2012, 03:01

mindmind wrote:

Please help me understand better :

The remainder of 4^40 would be same as that of 4 . Is my approach wrong? Can you provide similar examples, where it is not.

In this case, \(4^{40}\) is a \(M7+4\) : \(\,\,4^{40}=2^{80}\) and \(80=M3+2\) (the cycle is 3, because \(2^{3}=8=M7+1\)). \(4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4\).

Consider for example \(5^{40}\): \(5=7\cdot{0}+5=M7+5\) and \(5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6\). \(5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2\) and not \(M7+5\).[/quote]

Yes, Agreed So I should consider : Something near to Remainder 1

Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime eg : 4^40 divided by 7 give a remainder of 4.[/quote]

I am not sure what you mean here: Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime eg : 4^40 divided by 7 give a remainder of 4.

\(3^4=81=11\cdot{7}+4\). \(7\) is prime, but the remainder is neither \(3\), nor a prime. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Remainder problem. [#permalink]
17 Oct 2012, 03:09

EvaJager wrote:

mindmind wrote:

Please help me understand better :

The remainder of 4^40 would be same as that of 4 . Is my approach wrong? Can you provide similar examples, where it is not.

In this case, \(4^{40}\) is a \(M7+4\) : \(\,\,4^{40}=2^{80}\) and \(80=M3+2\) (the cycle is 3, because \(2^{3}=8=M7+1\)). \(4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4\).

Consider for example \(5^{40}\): \(5=7\cdot{0}+5=M7+5\) and \(5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6\). \(5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2\) and not \(M7+5\).

Yes, Agreed So I should consider : Something near to Remainder 1

Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime eg : 4^40 divided by 7 give a remainder of 4.[/quote]

I am not sure what you mean here: Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime eg : 4^40 divided by 7 give a remainder of 4.

\(3^4=81=11\cdot{7}+4\). \(7\) is prime, but the remainder is neither \(3\), nor a prime.[/quote]

Yes.. I was referring to 3.. and not the remainder 4.. But I got your point.. thanks..

Re: What is the remainder when you divide 2^200 by 7? [#permalink]
26 Oct 2012, 10:34

Another approach is the Theorem of Remainder 2^200 = 4*(2^3)^66;7 = 2^3 -1 Theorem of Remainder of f(X)/X-a is f(a) 2^200 = 4*f(X)/X-a X=2^3 a = 1 ==> Remainder of f(X)/X-a = 1 Remainder of 2^200 divided by7 is 4*1 = 4 Brother Karamazov

Re: What is the remainder when you divide 2^200 by 7? [#permalink]
08 Aug 2013, 22:39

Easiest and the smallest possible solution ever:

Keep your mind open while dealing with such question.It is not that you have to be math pro for that.

REM(2^200/7) [ REM(x/y) means remainder when x is divided by y]

We know that : Rem when 8 is divided by 7 is '1'.

Also by powers of 2 we can reach to 8.Using this concept:

[ (2^3)^198 * 2^2] /7

[ (8)^198 * 2^2] /7

Since REM(8/7) =1

We are left with REM(4/7) = 4 _________________

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

Re: What is the remainder when you divide 2^200 by 7? [#permalink]
12 Nov 2014, 15:55

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. _________________

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...

I started running back in 2005. I finally conquered what seemed impossible. Not sure when I would be able to do full marathon, but this will do for now...