Find all School-related info fast with the new School-Specific MBA Forum

It is currently 18 Sep 2014, 08:00

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

What is the remainder when you divide 2^200 by 7?

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 10 Oct 2012
Posts: 10
Followers: 0

Kudos [?]: 7 [0], given: 13

What is the remainder when you divide 2^200 by 7? [#permalink] New post 16 Oct 2012, 19:42
4
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

57% (01:38) correct 43% (00:42) wrong based on 123 sessions
What is the remainder when you divide 2^200 by 7?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

[Reveal] Spoiler:
my approach :

2^x has a cyclicity of 4
Therefore, Rem(200/4) = 0

Rem(2^0/7) =1

Am i missing something here?

OA is D
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Oct 2012, 03:22, edited 1 time in total.
Renamed the topic and edited the question.
3 KUDOS received
Intern
Intern
avatar
Joined: 23 May 2012
Posts: 31
Followers: 0

Kudos [?]: 9 [3] , given: 11

Re: Remainder problem. [#permalink] New post 16 Oct 2012, 20:18
3
This post received
KUDOS
2^{200} = (2^{5})^{40}

32 =28 +4

= (M7+4)^{40} ; M7 is a multiple of 7

= M7+4

so the remainder is 4.

Hope it helps.
Expert Post
2 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4772
Location: Pune, India
Followers: 1114

Kudos [?]: 5051 [2] , given: 164

Re: Remainder problem. [#permalink] New post 16 Oct 2012, 20:46
2
This post received
KUDOS
Expert's post
g3kr wrote:
I am confused . Please help!

What is the remainder when you divide 2^200 by 7?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

my approach :

2^x has a cyclicity of 4
Therefore, Rem(200/4) = 0

Rem(2^0/7) =1

Am i missing something here?

OA is D


I think you are getting confused between cyclicity of last digit and cyclicity of remainders.

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64

If you see, the last digits are 2, 4, 8, 6 i.e. cyclicity of 4.

On the other hand,

2^1/7 Rem = 2
2^2/7 Rem = 4
2^3/7 Rem = 1
2^4/7 Rem = 2
2^5 / 7 Rem = 4
2^6/7 Rem = 1

Here the cyclicity is 3.
2^{198} will give a remainder of 1. 2^{200} gives a remainder of 4.

Or, you can easily use binomial theorem here.
\frac{2^{200}}{7} = 2*2*\frac{2^{198}}{7} = 4*\frac{8^{66}}{7} = 4*\frac{(7 + 1)^{66}}{7}

Remainder must be 4. (Check out this link: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

1 KUDOS received
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 70

Kudos [?]: 515 [1] , given: 43

GMAT Tests User
Re: Remainder problem. [#permalink] New post 17 Oct 2012, 02:29
1
This post received
KUDOS
mindmind wrote:
EvaJager wrote:
mindmind wrote:
2^{200} = (2^{5})^{40}

32 =28 +4

= (M7+4)^{40} ; M7 is a multiple of 7

= M7+4

so the remainder is 4.

Hope it helps.


(M7+4)^{40}=M7+4^{40} then you have to find the remainder of 4^{40} when divided by 7.

Instead of taking 32 = 28 +4, look for a power of 2 which gives a remainder of 1 when divided by 7.
The smallest one is 8 = 7 + 1.

Therefore, 2^{200}=(2^3)^{66}\cdot{2^2}=8^{66}\cdot{4}=(7+1)^{66}\cdot{4}=(M7+1)\cdot4=M7+4.


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.
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.
Intern
Intern
avatar
Joined: 24 May 2012
Posts: 6
Concentration: Technology, Entrepreneurship
Followers: 0

Kudos [?]: 0 [0], given: 13

Re: Remainder problem. [#permalink] New post 16 Oct 2012, 20:11
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.
Intern
Intern
avatar
Joined: 23 May 2012
Posts: 31
Followers: 0

Kudos [?]: 9 [0], given: 11

Re: Remainder problem. [#permalink] New post 16 Oct 2012, 20:19
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 70

Kudos [?]: 515 [0], given: 43

GMAT Tests User
Re: Remainder problem. [#permalink] New post 17 Oct 2012, 00:17
mindmind wrote:
2^{200} = (2^{5})^{40}

32 =28 +4

= (M7+4)^{40} ; M7 is a multiple of 7

= M7+4

so the remainder is 4.

Hope it helps.


(M7+4)^{40}=M7+4^{40} then you have to find the remainder of 4^{40} when divided by 7.

Instead of taking 32 = 28 +4, look for a power of 2 which gives a remainder of 1 when divided by 7.
The smallest one is 8 = 7 + 1.

Therefore, 2^{200}=(2^3)^{66}\cdot{2^2}=8^{66}\cdot{4}=(7+1)^{66}\cdot{4}=(M7+1)\cdot4=M7+4.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Intern
Intern
avatar
Joined: 23 May 2012
Posts: 31
Followers: 0

Kudos [?]: 9 [0], given: 11

Re: Remainder problem. [#permalink] New post 17 Oct 2012, 00:29
EvaJager wrote:
mindmind wrote:
2^{200} = (2^{5})^{40}

32 =28 +4

= (M7+4)^{40} ; M7 is a multiple of 7

= M7+4

so the remainder is 4.

Hope it helps.


(M7+4)^{40}=M7+4^{40} then you have to find the remainder of 4^{40} when divided by 7.

Instead of taking 32 = 28 +4, look for a power of 2 which gives a remainder of 1 when divided by 7.
The smallest one is 8 = 7 + 1.

Therefore, 2^{200}=(2^3)^{66}\cdot{2^2}=8^{66}\cdot{4}=(7+1)^{66}\cdot{4}=(M7+1)\cdot4=M7+4.


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.
Intern
Intern
avatar
Joined: 23 May 2012
Posts: 31
Followers: 0

Kudos [?]: 9 [0], given: 11

Re: Remainder problem. [#permalink] New post 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.
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 70

Kudos [?]: 515 [0], given: 43

GMAT Tests User
Re: Remainder problem. [#permalink] New post 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.

Intern
Intern
avatar
Joined: 23 May 2012
Posts: 31
Followers: 0

Kudos [?]: 9 [0], given: 11

Re: Remainder problem. [#permalink] New post 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..
Intern
Intern
avatar
Joined: 11 Jul 2012
Posts: 46
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: What is the remainder when you divide 2^200 by 7? [#permalink] New post 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
Director
Director
avatar
Joined: 03 Aug 2012
Posts: 905
Concentration: General Management, General Management
Schools: ISB '16
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
Followers: 12

Kudos [?]: 204 [0], given: 312

Premium Member CAT Tests
Re: What is the remainder when you divide 2^200 by 7? [#permalink] New post 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
_____________________________________________________________________________

Senior Manager
Senior Manager
avatar
Joined: 10 Jul 2013
Posts: 343
Followers: 3

Kudos [?]: 112 [0], given: 102

Re: Remainder problem. [#permalink] New post 09 Aug 2013, 01:56
VeritasPrepKarishma wrote:
g3kr wrote:
I am confused . Please help!

What is the remainder when you divide 2^200 by 7?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

my approach :

2^x has a cyclicity of 4
Therefore, Rem(200/4) = 0

Rem(2^0/7) =1

Am i missing something here?

OA is D


I think you are getting confused between cyclicity of last digit and cyclicity of remainders.

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64

If you see, the last digits are 2, 4, 8, 6 i.e. cyclicity of 4.

On the other hand,

2^1/7 Rem = 2
2^2/7 Rem = 4
2^3/7 Rem = 1
2^4/7 Rem = 2
2^5 / 7 Rem = 4
2^6/7 Rem = 1

Here the cyclicity is 3.
2^{198} will give a remainder of 1. 2^{200} gives a remainder of 4.

Or, you can easily use binomial theorem here.
\frac{2^{200}}{7} = 2*2*\frac{2^{198}}{7} = 4*\frac{8^{66}}{7} = 4*\frac{(7 + 1)^{66}}{7}

Remainder must be 4. (Check out this link: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/)



...
2^200 = (2^3)^66 × 2^2 = (7+1)^66 × 4
1^66 × 4 = 4 (Answer)

The taste of your own medicine ???????
:-D :-D :-D :-D :-D
_________________

Asif vai.....

Re: Remainder problem.   [#permalink] 09 Aug 2013, 01:56
    Similar topics Author Replies Last post
Similar
Topics:
2 Experts publish their posts in the topic What is the remainder when 333^222 is divided by 7? jonyg 11 21 Jul 2013, 01:16
4 Experts publish their posts in the topic What is the remainder when you divide 2^200 by 7? carcass 7 02 Oct 2011, 10:50
What is the remainder when 7^n + 2 is divided by 5 (1) when vd 2 19 Jun 2008, 03:51
What is integer n? 1. When divided by 7, remainder is 3 2. vivek123 13 02 Apr 2006, 09:55
What is the remainder when 7^345 +7^11 -2 is divided by 7 joemama142000 6 20 Feb 2006, 22:30
Display posts from previous: Sort by

What is the remainder when you divide 2^200 by 7?

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.