Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k$?
Matthew Harrington 
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?
$\endgroup$ 23 Answers
$\begingroup$Since the cyclicity of $2$ is $2, 4, 8, 6$. It implies that for all $ n \in \mathbb{N}$.
$$2^{4n+1} \equiv 2 \mod 10 \\ 2^{4n+2} \equiv 4 \mod 10 \\ 2 ^{4n+3} \equiv 8 \mod 10 \\ 2^{4n+4} \equiv 6 \mod 10$$
Since $2008$ is of the form $4n$. So, $2^{2008} \equiv 6 \mod 10$.
But $2008 \equiv 8 \mod 10 \implies 2008^2 \equiv 8^2 \equiv 64 \equiv 4 \mod 10$.
So, $k = 2008^2 + 2^{2008} \equiv 4 + 6 \equiv 0 \mod 10 \implies k^2 \equiv 0 \mod 10$.
Now, $2^{2008} \equiv 0 \mod 4$ and $2008 \equiv 0 \mod 4 \implies 2008^2 \equiv 0 \mod 4$.
$$k = 2^{2008} + 2008^2 \equiv 0 + 0\equiv 0 \mod 4 $$
So, $k$ is of the form $4n$ for some positive integer $n$. So,
$$ 2^{k} \equiv 2^{4n} \equiv 6 \mod 10 $$
Therefore, $$2^{k} + k^{2} \equiv 6 + 0 \equiv 6 \mod 10$$
So, the last digit of $2^k + k^2$ is $6$.
$\endgroup$ 1 $\begingroup$$k$ is a huge even number, hence $k^2+2^k$ is for sure an even number. To compute the last digit of $k^2+2^k$, we just need to understand what $k^2+2^k\pmod{5}$ is. We have $k\equiv 0\pmod{4}$ and $$ k \equiv 3^2+4^{1004} \equiv -1+1\equiv 0\pmod{5}, $$ hence $k^2+2^k\equiv 0+1\equiv 1\pmod{5}$ by Fermat's little theorem, so the last digit of $k^2+2^k$ is $\color{red}{6}$.
$\endgroup$ $\begingroup$$$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 $6+0=6$.
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