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If k is a positive integer, What is the remainder when 2^k is divided [#permalink]
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23 Jan 2012, 21:55
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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: (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
Please help whether the above approach can be applied in the problem?
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Last edited by Bunuel on 20 Oct 2014, 08:35, edited 2 times in total.
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Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]
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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
Please help whether the above approach can be applied in the problem? 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. Answer: B. Hope it's clear.
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Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]
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Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]
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Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]
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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
Please help whether the above approach can be applied in the problem? Divisible not dividable bro Take it easy Cheers! J



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Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]
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09 Mar 2014, 13:21
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: (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
Please help whether the above approach can be applied in the problem? Divisible not dividable bro Take it easy Cheers! J ______________ Thank you. Edited.
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Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]
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20 Oct 2014, 08:27
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
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Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]
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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
Please help whether the above approach can be applied in the problem? 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. Answer: B. 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.



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Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]
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02 Nov 2014, 02:22
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: (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
Please help whether the above approach can be applied in the problem? 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.
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Re: If k is a positive integer, What is the remainder when 2^k is divided [#permalink]
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22 Mar 2016, 02:47
here the trick is to realise that the cylicity of 2 => Four hence statement 2 is sufficient and the remainder will be always => 6
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