TITLE: How to find $1 + 2^2 + \cdots + 2013^{2013} \equiv \; ? \pmod{10}$
QUESTION [8 upvotes]: Question:
let
$$a_{n}=1+2^2+3^3+\cdots+n^n$$
find this 
$$a_{2013}=1^2+2^2+\cdots+2013^{2013} \equiv \;    \pmod{10}$$
My try: maybe  $a_{n} \pmod{10}$ periodic, but I can't find it. Maybe this problem has nice methods. 
Thank you for you help!

REPLY [3 votes]: as $\phi(10) = 4$ then $n^k=n^r (mod 10)$ where $k=r$ mod $(4)$
example : $2=2 (mod 10), 2^2=4 (mod 10), 2^3= 8 (mod 10), 2^4= 6 (mod 10), 2^5= 2 (mod 10)$
          $4=4 (mod 10), 4^2=6 (mod 10), 4^3= 4 (mod 10), 4^4= 6 (mod 10)$..
and so on there is always a cycle of 4(not necessarily minimal) for any n that you pick..
now i will rewrite the sum as : 
$1 + \sum_{n=1}^{503} (4n-2)^{4n-2} +\sum_{n=1}^{503} (4n-1)^{4n-1}+\sum_{n=1}^{503} (4n)^{4n}+\sum_{n=1}^{503} (4n+1)^{4n+1}$.
observe that:
\begin{align}
(\sum_{n=1}^{503} (4n-2)^{4n-2}) (mod 10)=(\sum_{n=1}^{503} (4n-2)^{4n-2} (mod 10))=(\sum_{n=1}^{503} (4n-2)^{2} (mod 10))
\end{align}\begin{align}
(\sum_{n=1}^{503} (4n-1)^{4n-1}) (mod 10)=(\sum_{n=1}^{503} (4n-1)^{4n-1} (mod 10))=(\sum_{n=1}^{503} ((4n-1)^{3}(mod 10))\end{align}\begin{align}
(\sum_{n=1}^{503} (4n)^{4n}) (mod 10)=(\sum_{n=1}^{503} (4n)^{4n} (mod 10))=(\sum_{n=1}^{503} (4n)^{4} (mod 10))\end{align}\begin{align}
(\sum_{n=1}^{503} (4n+1)^{4n+1}) (mod 10)=(\sum_{n=1}^{503} (4n+1)^{4n+1} (mod 10))=(\sum_{n=1}^{503} (4n+1) (mod 10))
\end{align}
 the proof of some of the above work is left to the reader.
observe now that $n (mod 10) =(n+10) mod (10)$ therefore: all the sums will be reduced to:
\begin{align}
(\sum_{n=1}^{503} (4n-2)^{2} (mod 10))=(\sum_{n=501}^{503} (4n-2)^{2} (mod 10))=(\sum_{n=1}^{3} (4n-2)^{2} (mod 10))= 0 mod(10)\end{align}\begin{align}
(\sum_{n=1}^{503} ((4n-1)^{3}(mod 10))=(\sum_{n=501}^{503} ((4n-1)^{3}(mod 10))=(\sum_{n=1}^{3} ((4n-1)^{3}(mod 10))= 1 mod(10)\end{align}\begin{align}
(\sum_{n=1}^{503} (4n+1) (mod 10))=(\sum_{n=501}^{503} (4n+1) (mod 10))=(\sum_{n=1}^{3} (4n+1) (mod 10))= 7mod(10)\end{align}\begin{align}
(\sum_{n=1}^{503} (4n)^{4} (mod 10))=(\sum_{n=501}^{503} (4n)^{4} (mod 10))=(\sum_{n=1}^{3} (4n)^{4} (mod 10))=8mod(10)\end{align}
consequently :
$1 + 2^2 + \cdots + 2013^{2013} \equiv \; (1+0+1+7+8) \pmod{10}\equiv \; 7 \pmod{10}$