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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?

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

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.

(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?

(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?