It is currently 28 Jun 2017, 12:38

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

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

Hide Tags

Intern
Joined: 10 Oct 2012
Posts: 10
What is the remainder when you divide 2^200 by 7? [#permalink]

Show Tags

16 Oct 2012, 20:42
1
KUDOS
9
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

58% (01:47) correct 42% (00:48) wrong based on 329 sessions

HideShow timer Statistics

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, 04:22, edited 1 time in total.
Renamed the topic and edited the question.
Intern
Joined: 24 May 2012
Posts: 6
Concentration: Technology, Entrepreneurship

Show Tags

16 Oct 2012, 21:11
1
KUDOS
2
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.
Intern
Joined: 23 May 2012
Posts: 31

Show Tags

16 Oct 2012, 21:18
3
KUDOS
1
This post was
BOOKMARKED
$$2^{200} = (2^{5})^{40}$$

32 =28 +4

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

= M7+4

so the remainder is 4.

Hope it helps.
Intern
Joined: 23 May 2012
Posts: 31

Show Tags

16 Oct 2012, 21:19
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7449
Location: Pune, India

Show Tags

16 Oct 2012, 21:46
3
KUDOS
Expert's post
3
This post was
BOOKMARKED
g3kr wrote:

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

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Show Tags

17 Oct 2012, 01:17
1
This post was
BOOKMARKED
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
Joined: 23 May 2012
Posts: 31

Show Tags

17 Oct 2012, 01:29
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, $$2^{200}=(2^3)^{66}\cdot{2^2}=8^{66}\cdot{4}=(7+1)^{66}\cdot{4}=(M7+1)\cdot4=M7+4$$.

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.
Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Show Tags

17 Oct 2012, 03:29
1
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, $$2^{200}=(2^3)^{66}\cdot{2^2}=8^{66}\cdot{4}=(7+1)^{66}\cdot{4}=(M7+1)\cdot4=M7+4$$.

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, 03:48, edited 1 time in total.
Intern
Joined: 23 May 2012
Posts: 31

Show Tags

17 Oct 2012, 03:43

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
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Show Tags

17 Oct 2012, 04:01
mindmind wrote:

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
Joined: 23 May 2012
Posts: 31

Show Tags

17 Oct 2012, 04:09
EvaJager wrote:
mindmind wrote:

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
Joined: 11 Jul 2012
Posts: 46
Re: What is the remainder when you divide 2^200 by 7? [#permalink]

Show Tags

26 Oct 2012, 11:34
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
Joined: 03 Aug 2012
Posts: 894
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Re: What is the remainder when you divide 2^200 by 7? [#permalink]

Show Tags

08 Aug 2013, 23:39
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
Joined: 10 Jul 2013
Posts: 334

Show Tags

09 Aug 2013, 02:56
VeritasPrepKarishma wrote:
g3kr wrote:

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 ???????

_________________

Asif vai.....

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16024
Re: What is the remainder when you divide 2^200 by 7? [#permalink]

Show Tags

12 Nov 2014, 16:55
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16024
Re: What is the remainder when you divide 2^200 by 7? [#permalink]

Show Tags

25 Jan 2016, 03:16
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.
_________________
BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2185
Re: What is the remainder when you divide 2^200 by 7? [#permalink]

Show Tags

21 Mar 2016, 09:37
We can use the binomial approach here => 4*2^198 => 4* (7P+1) => 7Q+4 => remainder => 4
_________________

Give me a hell yeah ...!!!!!

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16024
Re: What is the remainder when you divide 2^200 by 7? [#permalink]

Show Tags

03 Apr 2017, 10:45
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 you divide 2^200 by 7?   [#permalink] 03 Apr 2017, 10:45
Similar topics Replies Last post
Similar
Topics:
6 What is the remainder when 5^68 is divided by 7? 8 15 May 2017, 10:21
3 What is the remainder when 7^442 is divided by 10? 5 24 Sep 2016, 22:14
1 If n divided by 7 has a remainder of 2, what is the remainder when 3 8 03 Jun 2017, 10:28
34 What is the remainder when 333^222 is divided by 7? 19 02 Dec 2016, 03:32
10 What is the remainder when you divide 2^200 by 7? 8 08 Dec 2015, 12:16
Display posts from previous: Sort by