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 dividable 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 dividable 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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60% (02:04) correct
