Last Digit

We have powers of $2$ , $3$ and $7$ , so we analyze their last digits: $\begin{array} { c | c c c c } d & d^2 & d^3 & d ^ 4 & d^ 5\\ \hline 2 & 4 & 8 & 6 & 2 \\ 3 & 9 & 7 & 1 & 3 \\ 7 & 9 & 3 & 1 & 7 \\ \end{array}$ All repeat in a cycle of four, so we use $\pmod {4}$ for the powers.

Let $n=2^{2013}=(2^2)^{1006}\times2\equiv0\times2\equiv0\pmod{4}$

For $2^n$ , $n\equiv0\pmod4\iff2^n\equiv\ 6\pmod{10}$ For $3^{n+1}$ , $n+1\equiv1\pmod4\iff3^{n+1}\equiv\ 3\pmod{10}$ For $7^{n+2}$ , $n+2\equiv2\pmod4\iff7^{n+2}\equiv\ 9\pmod{10}$ Then, $6+3+9=18\equiv\boxed{8\pmod{10}}$