We can find the last (units) digit by looking at the units digits of the four parts. We can quickly note that the units digits of $$0^{2016}$$ and $$1^{2016}$$ and $$6^{2016}$$are $$0$$, $$1$$ and$$6$$ respectively. Now then we have to look at powers of $$2:{{2}^{1}}=2$$, $${{2}^{2}}=4$$, $${{2}^{3}}=8$$, $${{2}^{4}}=16$$, and after this the units digits repeat ($${{2}^{5}}=3\underline{2}$$, $${{2}^{6}}=6\underline{4}$$, $${{2}^{7}}=12\underline{8}$$, $${{2}^{8}}=25\underline{6}$$, $$\cdots $$). We notice that when the indices are multiples of 4 (eg. $${{2}^{4}}=16$$, $${{2}^{8}}=256$$, $$\cdots $$) the units digit of the power of $$2$$ is $$6$$. Hence the units digit of $$2^{2016}+0^{2016}+1^{2016}+6^{2016}$$ is the same as the units digit of $$6 + 0 + 1 + 6$$, that is $$3$$. 
If you were to work out the answer to the sum  $$2^{2016}+0^{2016}+1^{2016}+6^{2016}$$  you would get a number with $$1569$$ digits, starting with $$566$$ $$136$$ $$001$$ $$\cdots$$  What is the last digit of this number? 

a. $$1$$ 
b. $$3$$ 
c. $$5$$ 
d. $$7$$ 
e. $$9$$ 