What is the remainder when 2^n is divided by 10?

I think the question should be \(2^n\).
Then from statement 1 n can take values 2,4,6,8... Then \(2^n\) will not give a unique remainder. Hence 1 is insufficient.
Form statement 2 n can take values 4,8,12,16... Then \(2^n\) will be \(2^4\), \(2^8\), \(2^{12}\)...(Cyclicity of 2 is 4) In all such cases, the remainder will be 6. Hence, 2 is sufficient.
B is the answer.

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The remainder when \(2^n\) is divided by \(10\) is the units digit of \(2^n\).

Now, \(2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16\), and \(2^5 = 32\).
So, the units digits of \(2^n\) have period \(4\):
They form the cycle \(2 -> 4 -> 8 -> 6\).
Thus, \(2^n\) has the units digit of \(6\) when \(n\) is a multiple of \(4\).
Condition 2) is sufficient.

Condition 1)
If \(n = 4\), then \(2^n = 2^4 = 16\) has the units digit of \(6\).
If \(n = 2\), then \(2^n = 2^2 = 4\) has the units digit of \(4\).
Since we don’t have a unique solution, condition 1) is not sufficient.

Therefore, B is the answer.

Answer: B

The powers for 2 follow a cyclic behavior (not sure if that's the correct word) - so every 2^4n will end with a 6 in the end. Making statement 2 sufficient.

Asked: What is the remainder when \(2^n\) is divided by \(10\)?

1) \(n\) is a positive multiple of \(2\)
n = 2k
2^n = 2^{2k}
The remainder when 2^n is divided by 10 = {4,6}
NOT SUFFICIENT

2) \(n\) is a positive multiple of \(4\)
n = 4k
2^n = 2^{4k}
The remainder when 2^n is divided by 10 = 6
SUFFICIENT

IMO B