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# If k is a positive integer, what is the reminder when 2^k...

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If k is a positive integer, what is the reminder when 2^k... [#permalink]  23 Jan 2012, 20:55
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Difficulty:

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Question Stats:

57% (02:17) correct 43% (01:13) wrong based on 263 sessions
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

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Last edited by Bunuel on 09 Mar 2014, 12:20, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
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Kudos [?]: 26217 [0], given: 2706

Re: If k is a positive integer, What is the reminder when 2^k... [#permalink]  24 Jan 2012, 01:05
Expert's post
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.
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Kudos [?]: 230 [0], given: 32

Re: If k is a positive integer, what is the reminder when 2^k... [#permalink]  25 Jan 2012, 08:57
Understand. Thank u bunnel.
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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

Math Expert
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Posts: 29642
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Kudos [?]: 26217 [0], given: 2706

Re: If k is a positive integer, what is the reminder when 2^k... [#permalink]  06 Jun 2013, 05:10
Expert's post
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Kudos [?]: 164 [1] , given: 268

Re: If k is a positive integer, what is the reminder when 2^k... [#permalink]  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: 29642
Followers: 3488

Kudos [?]: 26217 [0], given: 2706

Re: If k is a positive integer, what is the reminder when 2^k... [#permalink]  09 Mar 2014, 12:21
Expert's post
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

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Thank you. Edited.
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Re: If k is a positive integer, what is the reminder when 2^k...   [#permalink] 09 Mar 2014, 12:21
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