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(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?

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.

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

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20 Oct 2014, 09:31

1

This post received 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

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.

(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.

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

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25 Apr 2017, 08:32

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