Find the unit digit of (abc)^k

Statement 1: \(c\) is prime and even. \(c = 2\). I don't know anything about \(a, b, or\) \(k\). Insufficient.

Statement 2: \(k\) leaves a remainder of \(1\) when divided by \(4\). Possible options for \(k\) are \(0, 5, 9, 13, ...\)I don't know anything about \(a, b, or\) \(c\). Insufficient.

Statement 1& 2:
Combining yields
\((ab2)^1\) and the units digit here is \(2\)

\((ab2)^5\) and the units digit here is \(2\) (\(2*2*2*2*2\) \(= 3\)2)

\((ab2)^9\) and the units digit here is \(2\) (I'm not writing this one all the way out. \(2^9 = 51\)2)

I'm not going to test \((ab2)^{13}\) since I see that I have three answers where the unit digit is \(2\).

The question asks me to find the units digit of \((abc)^k\) and in all of the cases it is \(2\). Therefore C is sufficient.
Answer: C