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

It is currently 12 Jul 2014, 13:05

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

If n^m leaves a remainder of 1 after division by 7 for all p

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Manager
Manager
User avatar
Joined: 11 Feb 2011
Posts: 147
Followers: 3

Kudos [?]: 21 [1] , given: 21

GMAT Tests User
If n^m leaves a remainder of 1 after division by 7 for all p [#permalink] New post 19 Jun 2011, 09:42
1
This post received
KUDOS
1
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  55% (medium)

Question Stats:

41% (02:39) correct 58% (01:11) wrong based on 119 sessions
If n^m leaves a remainder of 1 after division by 7 for all positive integers n that are not multiples of 7, then m could be equal to :

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
[Reveal] Spoiler: OA

_________________

target:-810 out of 800!

Manager
Manager
User avatar
Joined: 07 Oct 2010
Posts: 181
Followers: 5

Kudos [?]: 89 [0], given: 10

GMAT Tests User
Re: tough 1 [#permalink] New post 19 Jun 2011, 10:16
We need to find m for all positive integers.
Therefore, if we take a case of 2 then 2^3 and 2^6 will leave remainder of 1 after dividing by 7. we will restrict to 6 and not more than that because it is a largest value in the options and we have to select only one of them.

Then, i took a case of 3 ...In that 3^6 leaves remainder of 1 when divided by 7.

Now, we can conclude that the value of m will be 6 because it common value in both cases and one of the given option has to be correct. Hence m=6
thus E
_________________

Check this link

http://www.rupeemail.com/rupeemail/invite.do?in=NzYxOTkxJSMlRzF0NGJKQnllT1kzVmRmbTVBb2tLZ1Q4RA==

http://www.rupeemail.com/rupeemail/pages/rupeemail/images/joinImage.gif

4 KUDOS received
Current Student
avatar
Joined: 26 May 2005
Posts: 571
Followers: 18

Kudos [?]: 98 [4] , given: 13

GMAT Tests User
Re: tough 1 [#permalink] New post 19 Jun 2011, 10:24
4
This post received
KUDOS
AnkitK wrote:
if n^m leaves a remainder of 1 after division by 7 for all positive integers n that are not multiples of 7, then m could be equal to :

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

What is the underlying concept ?


its E
pick up the numbers

ex n = 2
2^1/7 = 2
2^2/7 = 4
2^3/7 = 1
2^4/7 = 4
2^5/7 = 4
2^6 /7 = 1

so m can be 3 or 6
now take n = 3
3^3/7 = 6
3^6/7 = 1

hence 6.

concept:
we know that the remainder when an interger is divided by 7 are 1,2,3,4,5,6 (n-1 concept)
now suppose m =2
the remainder will also be raised by m
1,4,9,16,25,36
divide this by 7 we dont get 1 consistently

try for 3,4,5,.. once u get 6 u will get all 1s
1,64,729,4096, 15625 , 46656) all these number when divided by 7 leaves a remainder 1.

hope this helps.
Manager
Manager
User avatar
Joined: 11 Feb 2011
Posts: 147
Followers: 3

Kudos [?]: 21 [0], given: 21

GMAT Tests User
Re: tough 1 [#permalink] New post 19 Jun 2011, 10:26
Nice explanation Sudhir.
_________________

target:-810 out of 800!

Director
Director
avatar
Joined: 03 Aug 2012
Posts: 865
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GPA: 3.7
Followers: 11

Kudos [?]: 145 [0], given: 294

Premium Member CAT Tests
Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink] New post 09 Aug 2013, 04:07
Somehow didn't get this question as per below.

n^m = 7Q + 1

n != 7A (where != is not equal to)

Consider n=6 and answer option (A) m=2

6^2 = 36 = 7*5 + 1

Why not (A) then?

Rgds,
TGC!
_________________

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

Intern
Intern
avatar
Joined: 02 Jun 2013
Posts: 20
Followers: 0

Kudos [?]: 1 [0], given: 70

Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink] New post 09 Aug 2013, 04:13
hi everyone,

I solved this question like this , as 7 is a prime no and n can only take values which are not multiples of 7 so n and 7 will be co prime hence as per fermat little theorem value of m will be 7-1 = 6 for any value of n which are not multiples of n
Intern
Intern
avatar
Joined: 21 Aug 2013
Posts: 10
Followers: 0

Kudos [?]: 0 [0], given: 169

Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink] New post 08 Sep 2013, 07:55
abhishekkhosla wrote:
hi everyone,

I solved this question like this , as 7 is a prime no and n can only take values which are not multiples of 7 so n and 7 will be co prime hence as per fermat little theorem value of m will be 7-1 = 6 for any value of n which are not multiples of n


for those who know fermat little theorem(s) its good but for others, plugging in will always help as explained by Sudhir... thanks Abhishek though for reminding this theorem
Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4519
Location: Pune, India
Followers: 1013

Kudos [?]: 4323 [1] , given: 161

Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink] New post 17 Sep 2013, 21:03
1
This post received
KUDOS
Expert's post
AnkitK wrote:
If n^m leaves a remainder of 1 after division by 7 for all positive integers n that are not multiples of 7, then m could be equal to :

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6


Responding to a pm:

It is a conceptual question and it easy to figure out if you understand binomial theorem discussed here: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

n^m leaves a remainder of 1 after division by 7 for all n (obviously n cannot be a multiple of 7 because that would leave a remainder of 0)
So n can be of the form (7a + 1) or (7a + 2) or (7a + 3) or (7a + 4) or (7a +5) or (7a + 6)

According to binomial, the remainder I will get when I divide n^m by 7 will depend on the last term i.e. 1^m or 2^m or 3^m or 4^m or 5^m or 6^m.
We need a value of m such that when we divide any one of these 6 terms by 7, we always get a remainder 1.

Can m be 2? 1^2 leaves remainder 1 but 2^2 leaves remainder 4. So no
Can m be 3? 1^3 leaves remainder 1, 2^3 leaves remainder 1. 3^3 leaves remainder 6. So no
Can m be 4? 1^4 leaves remainder 1 but 2^4 leaves remainder 2. So no
Can m be 5? 1^5 leaves remainder 1 but 2^5 leaves remainder 4. So no'

m must be 6 because that is the only option left.
1^6 leaves remainder 1, 2^6 leaves remainder 1 etc. So we see that we are correct.
_________________

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

Senior Manager
Senior Manager
avatar
Joined: 17 Apr 2013
Posts: 488
Concentration: Entrepreneurship, Leadership
Schools: HBS '16
GMAT Date: 11-30-2013
GPA: 3.3
Followers: 1

Kudos [?]: 27 [0], given: 230

Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink] New post 18 Sep 2013, 01:14
VeritasPrepKarishma wrote:
AnkitK wrote:
If n^m leaves a remainder of 1 after division by 7 for all positive integers n that are not multiples of 7, then m could be equal to :

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6


Responding to a pm:

It is a conceptual question and it easy to figure out if you understand binomial theorem discussed here: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

n^m leaves a remainder of 1 after division by 7 for all n (obviously n cannot be a multiple of 7 because that would leave a remainder of 0)
So n can be of the form (7a + 1) or (7a + 2) or (7a + 3) or (7a + 4) or (7a +5) or (7a + 6)

According to binomial, the remainder I will get when I divide n^m by 7 will depend on the last term i.e. 1^m or 2^m or 3^m or 4^m or 5^m or 6^m.
We need a value of m such that when we divide any one of these 6 terms by 7, we always get a remainder 1.

Can m be 2? 1^2 leaves remainder 1 but 2^2 leaves remainder 4. So no
Can m be 3? 1^3 leaves remainder 1, 2^3 leaves remainder 1. 3^3 leaves remainder 6. So no
Can m be 4? 1^4 leaves remainder 1 but 2^4 leaves remainder 2. So no
Can m be 5? 1^5 leaves remainder 1 but 2^5 leaves remainder 4. So no'

m must be 6 because that is the only option left.
1^6 leaves remainder 1, 2^6 leaves remainder 1 etc. So we see that we are correct.


I understand your solution, But how is it associated with Binomial theorem, we have solved it with plugin here?
_________________

Like my post Send me a Kudos :) It is a Good manner.

Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4519
Location: Pune, India
Followers: 1013

Kudos [?]: 4323 [0], given: 161

Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink] New post 18 Sep 2013, 06:30
Expert's post
honchos wrote:
VeritasPrepKarishma wrote:
AnkitK wrote:
If n^m leaves a remainder of 1 after division by 7 for all positive integers n that are not multiples of 7, then m could be equal to :

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6


Responding to a pm:

It is a conceptual question and it easy to figure out if you understand binomial theorem discussed here: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

n^m leaves a remainder of 1 after division by 7 for all n (obviously n cannot be a multiple of 7 because that would leave a remainder of 0)
So n can be of the form (7a + 1) or (7a + 2) or (7a + 3) or (7a + 4) or (7a +5) or (7a + 6)

According to binomial, the remainder I will get when I divide n^m by 7 will depend on the last term i.e. 1^m or 2^m or 3^m or 4^m or 5^m or 6^m.
We need a value of m such that when we divide any one of these 6 terms by 7, we always get a remainder 1.

Can m be 2? 1^2 leaves remainder 1 but 2^2 leaves remainder 4. So no
Can m be 3? 1^3 leaves remainder 1, 2^3 leaves remainder 1. 3^3 leaves remainder 6. So no
Can m be 4? 1^4 leaves remainder 1 but 2^4 leaves remainder 2. So no
Can m be 5? 1^5 leaves remainder 1 but 2^5 leaves remainder 4. So no'

m must be 6 because that is the only option left.
1^6 leaves remainder 1, 2^6 leaves remainder 1 etc. So we see that we are correct.


I understand your solution, But how is it associated with Binomial theorem, we have solved it with plugin here?


The remainder you get when you divide (7a + 1)^m by 7 will be 1. The remainder you get when you divide (7a + 2)^m by 7 is determined by 2^m. This is determined by binomial theorem. The link explains you why.
_________________

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

Re: If n^m leaves a remainder of 1 after division by 7 for all p   [#permalink] 18 Sep 2013, 06:30
    Similar topics Author Replies Last post
Similar
Topics:
2 Experts publish their posts in the topic What is the remainder of the division 2^56 by 7? pbull78 4 31 Jan 2012, 22:05
5 Experts publish their posts in the topic What is the remainder, after division by 100, of 7^10 ? kkalyan 12 07 Oct 2011, 20:45
90 Experts publish their posts in the topic Positive integer n leaves a remainder of 4 after division by bchekuri 32 05 May 2010, 22:49
12 Experts publish their posts in the topic What is the sum of all 3 digit numbers that leave a remainde joyseychow 14 06 Aug 2009, 19:33
Experts publish their posts in the topic (P/S) Remainder 1 pretttyune 5 26 Nov 2007, 04:16
Display posts from previous: Sort by

If n^m leaves a remainder of 1 after division by 7 for all p

  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®.