Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k$? Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k.$
I thought of this
$$2008^2+2^{2008}\pmod{10} ≡ {-2}^2+{2^4}^{502}\pmod{10} ≡ 4+{-4}^{502}\pmod{10} ≡ 4+6^{251} \pmod{10}$$
but I still cannot prove it. Maybe there is a clever solution but so far I have been unable to spot it. Can anyone help me?

$$2008^2=2000^2+2*2000*8+8^2=>4 (mod 10)$$
$2^m=6 (mod 10) $ (m = $2008$)
$$4+6=0 (mod 10)$$
$$ k=0 (mod 10)$$
$$ k^2=0 (mod 10)$$
So, it is obvious that last digit of $2^k$ is $2^4=6 (mod 10)$ and last digit of $k^2$ is $0$.
So the answer is $6+0=6$.
