What is the remainder when 6^x is divided by 10

1 alone is sufficient. If X is 0, then the remainder is 1. In all other cases the remainder is 6

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Answer should be D.

Statement 1: x is a positive interger. Meaning X > 0 and is an Integer.

Now, \((6)^P\) (where P is a positive Integer) will always give the last digit as 6 and hence remainder will always remain 6. --> SUFFICIENT

Statement 2: This is just a filtered part of Statement 1. Meaning in Statement 1, I already said we will get a single solution for X > 0 and X as an integer. Now, this includes both even and Odd. This, Statement 2 is also sufficient.

Hence, D.
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6 raised to any positive integer power will Always give an integer that ends in '6' (meaning the unit place digit of that number would be 6). So such an integer, when divided by 10, would always give a remainder 6.

Here, each statement alone suffices to tell us that last digit of 6^x would be 6, hence remainder would be 6.

Hence D answer

Yeah true. My bad.
I jumped to the conclusion after reading statement 1.

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How does this change if it's not given x is an integer. Only positive.