If k is a positive integer, What is the remainder when 2^k is divided : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 16:49

### 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

# If k is a positive integer, What is the remainder when 2^k is divided

Author Message
TAGS:

### Hide Tags

Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 651
Followers: 42

Kudos [?]: 863 [1] , given: 39

If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

23 Jan 2012, 20:55
1
KUDOS
10
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

60% (02:14) correct 40% (01:19) wrong based on 567 sessions

### HideShow timer Statistics

If k is a positive integer, What is the remainder when 2^k is divided by 10?

(1) k is divisible by 10
(2) k is divisible by 4

My approach is as follows:
[Reveal] Spoiler:
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

[Reveal] Spoiler: OA

_________________

Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

Last edited by Bunuel on 20 Oct 2014, 07:35, edited 2 times in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93280 [2] , given: 10555

Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

24 Jan 2012, 01:05
2
KUDOS
Expert's post
4
This post was
BOOKMARKED
Baten80 wrote:
If k is a positive integer, What is the remainder when 2^k is divided by 10?
1) k is dividable by 10
2) k is dividable by 4

My approach is as follows:
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

General approach is correct, though the red parts are not.

The last digit of 2^k repeats in pattern of 4 (cyclicity is 4):
2^1=2 --> last digit is 2;
2^2=4 --> last digit is 4;
2^3=8 --> last digit is 8;
2^4=16 --> last digit is 6;

2^5=32 --> last digit is 2 again;

Now, when k itself is a multiple of 4 (when there is no remainder upon division k by cyclicity number), then the last digit will be the last digit of 2^4 (4th in pattern), so 6 not 1 (taking 2^0) as you've written.

If k is a positive integer, what is the remainder when 2^k is divided by 10?

Notice that all we need to know to answer the question is the last digit of 2^k.

(1) k is divisible by 10 --> different multiples of 10 yield different remainders upon division by 4 (for example 10/4 yields 2 and 20/4 yields 0), thus we can not get the single numerical value of the last digit of 2^k. Not sufficient.

(2) k is divisible by 4 --> as discussed, when k is a multiple of 4, the last digit of 2^k equals to the last digit of 2^4, which is 6. Integer ending with 6 yields remainder of 6 upon division by 10. Sufficient.

Hope it's clear.
_________________
Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 651
Followers: 42

Kudos [?]: 863 [0], given: 39

Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

25 Jan 2012, 08:57
Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93280 [0], given: 10555

Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

06 Jun 2013, 05:10
Expert's post
1
This post was
BOOKMARKED
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on remainders problems: remainders-144665.html

All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199

_________________
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 62

Kudos [?]: 594 [1] , given: 355

Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

17 Dec 2013, 09:32
1
KUDOS
Baten80 wrote:
If k is a positive integer, What is the remainder when 2^k is divided by 10?

(1) k is dividable by 10
(2) k is dividable by 4

My approach is as follows:
[Reveal] Spoiler:
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

Divisible not dividable bro

Take it easy
Cheers!
J
Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93280 [0], given: 10555

Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

09 Mar 2014, 12:21
jlgdr wrote:
Baten80 wrote:
If k is a positive integer, What is the remainder when 2^k is divided by 10?

(1) k is dividable by 10
(2) k is dividable by 4

My approach is as follows:
[Reveal] Spoiler:
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

Divisible not dividable bro

Take it easy
Cheers!
J

______________
Thank you. Edited.
_________________
Current Student
Joined: 02 Jul 2012
Posts: 215
Location: India
Schools: IIMC (A)
GMAT 1: 720 Q50 V38
GPA: 2.6
WE: Information Technology (Consulting)
Followers: 15

Kudos [?]: 224 [0], given: 84

Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

20 Oct 2014, 07:27
Baten80 wrote:
If k is a positive integer, What is the remainder when 2^k is divided by 10?

(1) k is divisible by 10
(2) k is divisible by 4

2^k divided by 10. The cycliicity of 2 when divided by 10 is 4.

1 - k is divisible by 10 - the number can be 10 (2) or 20(0) - Not Sufficient
2 - k is divisible by 4 - Sufficient.

Ans. B
_________________

Give KUDOS if the post helps you...

Manager
Joined: 26 May 2013
Posts: 93
Followers: 0

Kudos [?]: 29 [1] , given: 29

Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

20 Oct 2014, 08:31
1
KUDOS
Bunuel wrote:
Baten80 wrote:
If k is a positive integer, What is the remainder when 2^k is divided by 10?
1) k is dividable by 10
2) k is dividable by 4

My approach is as follows:
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

General approach is correct, though the red parts are not.

The last digit of 2^k repeats in pattern of 4 (cyclicity is 4):
2^1=2 --> last digit is 2;
2^2=4 --> last digit is 4;
2^3=8 --> last digit is 8;
2^4=16 --> last digit is 6;

2^5=32 --> last digit is 2 again;

Now, when k itself is a multiple of 4 (when there is no remainder upon division k by cyclicity number), then the last digit will be the last digit of 2^4 (4th in pattern), so 6 not 1 (taking 2^0) as you've written.

If k is a positive integer, what is the remainder when 2^k is divided by 10?

Notice that all we need to know to answer the question is the last digit of 2^k.

(1) k is divisible by 10 --> different multiples of 10 yield different remainders upon division by 4 (for example 10/4 yields 2 and 20/4 yields 0), thus we can not get the single numerical value of the last digit of 2^k. Not sufficient.

(2) k is divisible by 4 --> as discussed, when k is a multiple of 4, the last digit of 2^k equals to the last digit of 2^4, which is 6. Integer ending with 6 yields remainder of 6 upon division by 10. Sufficient.

Hope it's clear.

To add some clarity for myself and viewers:

Since the last digit in 2^k repeats in cycles of 4, you will ALWAYS know the last digit (and remainder) if k is a multiple of 4.

Therefore 2^4, 2^8,2^12. 2_16, etc.... will always have a last digit of 6.

If k is a multiple of 10, you know if k = 10, the last digit will be 4, and if k=20 the last digit will be 6, k=30 the last digit will be 4, etc... in repeating pattern. However without knowing the exact value of k you won't know the remainder.
Manager
Joined: 22 Jan 2014
Posts: 138
WE: Project Management (Computer Hardware)
Followers: 0

Kudos [?]: 54 [0], given: 135

Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

02 Nov 2014, 01:22
Baten80 wrote:
If k is a positive integer, What is the remainder when 2^k is divided by 10?

(1) k is divisible by 10
(2) k is divisible by 4

My approach is as follows:
[Reveal] Spoiler:
(1) k could be 10, 20, 30...
case i. if k = 10, 2^10, the cyclicity of 2 is 4 (10/4 = reminder 2) so 2^2 is divided by 10 and reminder is 4
case ii. if k = 20, 2^20, the cyclicity of 2 is 4 (20/4 = 5, 5/4 = reminder 1) so 2^1 is divided by 10 and reminder is 2
Insufficient.

(2) k = 4, 8, 12
2^4, the cyclicity of 2 is 4 (4/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
2^8, the cyclicity of 2 is 4 (8/4 = reminder 0) so 2^0 is divided by 10 and reminder is 1
Sufficient.

Ans. B

remainder by 10 means units digit.

1) k is div by 10
k = 10 ; 2^10 ends in 4
k = 20 ; 2^20 ends in 6
insufficient.

2) k is div by 4
2^(4k) always ends in 6
sufficient.

B.
_________________

Illegitimi non carborundum.

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13468
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

05 Mar 2016, 16:44
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: 1901
Followers: 49

Kudos [?]: 366 [0], given: 454

Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]

### Show Tags

22 Mar 2016, 01:47
here the trick is to realise that the cylicity of 2 => Four
hence statement 2 is sufficient and the remainder will be always => 6
_________________

Mock Test -1 (Integer Properties Basic Quiz) ---> http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1676182

Mock Test -2 (Integer Properties Advanced Quiz) --->http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1765951

Mock Test -2 (Evensand Odds Basic Quiz) --->http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1768023

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

Re: If k is a positive integer, What is the remainder when 2^k is divided   [#permalink] 22 Mar 2016, 01:47
Similar topics Replies Last post
Similar
Topics:
1 What is the remainder when 2^k is divided by 10? 3 12 Jan 2017, 17:16
29 What is the remainder when the positive integer n is divided 17 10 May 2009, 20:09
14 What is the remainder when the positive integer x is divided 9 07 Oct 2008, 05:25
whats the remainder when 2^k divided by 10 I. k dividable by 3 02 Jun 2007, 04:52
14 What is the remainder when the positive integer x is divided 12 15 Apr 2007, 04:57
Display posts from previous: Sort by

# If k is a positive integer, What is the remainder when 2^k is divided

 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®.