If k is a positive integer, what is the remainder...

The problem is asking for the remainder when 2^k is divided by 10. You need to realize that the problem lies with the last digit of 2^k, since the remainder of dividing a number by 10 will always be its last digit.
You should know that there is a certain regularity to the last digit, depending on the nature of k.
Let's start with the basics:
2^1 = 2 - last digit 2
2^2 = 4 - last digit 4
2^3 = 8 - last digit 8
2^4 = 16 - last digit 6
2^5 = 32 - last digit 2 - and the last digit repeats from now on.
This forms a pattern for 2^k:
a. when k = 4n + 1, last digit = 2
b. when k = 4n + 2, last digit = 4
c. when k = 4n + 3, last digit = 8
d. when k = 4n, last digit = 6.

This means that the last digit of 2^k is linked to whether k is divisible or not by 4, thus making the second stmt sufficient, with answer B

IMO B . Whats the OA

i dnt get it.... if the last digit keeps on changing then the remainder wil also be changing..tht means we dont get a specific value of the remainder then how come B is correct?

B) k is divisible by 4
=> k=4, 8, 12, 16....... we cant take 0 as k is positive

if k=4, 2^4=16. 16/10 =>remainder 6 (the last digit)
if k=8, 2^8= (2^4)*(2^4)=16*16= 256. 256/10 => remainder=6

if k=16, (2^4)*(2^4)*(2^4)*(2^4)=16^4 leaves last digit as 6. (6 multiplied by itself always leaves last digit as 6). 16^4/10 therefore leaves remainder 6.

B always gives a remainder of 6

Couple of things I have noticed with the GMAT. They like to ask a lot of questions with exponent's last digits repeating in patterns. For the power of 2 it repeats in the pattern: 2,4,6,8. Hence, if k is divisible by 10, you are not sure where you will be in the sequence. However, if k is divisible by 4, it does not matter how big k gets because the pattern consists of 4 numbers. Hence you will always be able to find the remainder.

You guys think this is sound reasoning? Just curious.

Couple of things I have noticed with the GMAT. They like to ask a lot of questions with exponent's last digits repeating in patterns. For the power of 2 it repeats in the pattern: 2,4,6,8. Hence, if k is divisible by 10, you are not sure where you will be in the sequence. However, if k is divisible by 4, it does not matter how big k gets because the pattern consists of 4 numbers. Hence you will always be able to find the remainder.

You guys think this is sound reasoning? Just curious.