What is the remainder of $1^1$+$2^2$+$3^3$+...+$2020^{2020}$ if divided by 10? (Without Calculator)

Question

Is there any way to calculate the remainder of $1^1$ + $2^2$ + $3^3$ + ... + $2020^{2020}$ when divided by 10 without calculator?

Answer 1

Hint $\bmod 10\!:\ n^{\large\color{#c00}{k+4j}}\equiv n^{\large k}\,$ for $\,k>0,\,$ by $\, n^{\large k+4j}\!-n^{\large k} \equiv n^{\large k}(n^{\large 4j}\!-1) \equiv 0\, $ by here.

So $\,(n\!+\!20)^{\large n+20}\!\equiv n^{\large\color{#c00}{ n+4\cdot 5}}\!\equiv n^{\large n}\,$ for $\,n>0\,$ so $\!\bmod 10$ the summands repeat in cycles of length $20,\,$ so we need only compute a single cycle sum, which is easy using above, e.g. $\,17^{\large 17}\!\equiv 7^{\large\color{#c00}{ 1+4\cdot 4}}\!\equiv 7^{\large 1}$.

Answer 2

You may split it into $\mod 2$ and $\mod 5$:

$\mod 2$: \begin{eqnarray*} \sum_{n=1}^{2020}n^n & \equiv_2 & 1010\cdot 1 \\ & \equiv_2 & 0 \\ \end{eqnarray*} $\mod 5$: \begin{eqnarray*} \sum_{n=1}^{2020}n^n & \equiv_5 & \sum_{k=0}^{403}\left((5k+1)^{5k+1} + (5k+2)^{5k+2}+(5k+3)^{5k+3}+(5k+4)^{5k+4} \right) \\ & \stackrel{\mbox{Fermat}}{\equiv_5} & \sum_{k=0}^{403}\left(1 + 2^{k+2}+3^{k+3}+4^{k+4} \right)\\ & \equiv_5 & \sum_{k=0}^{403}\left(1 + 2^{k+2}+(-2)^{k+3}+(-1)^{k+4} \right)\\ & \equiv_5 & -1 + \sum_{k=0}^{403}\left(2^{k+2}+(-2)^{k+3}\right) + 0\\ & \equiv_5 & -1 - \sum_{k=0}^{403}2^{k+2} \\ & \equiv_5 & -1 + \sum_{k=0}^{403}2^{k} \\ & \equiv_5 & -1 + 2^{404} - 1 \\ & \stackrel{\mbox{Fermat}}{\equiv_5} & -1 \\ \end{eqnarray*} So, $\sum_{n=1}^{2020}n^n \equiv_2 0$ and $\sum_{n=1}^{2020}n^n\equiv_5 -1 \equiv_5 4 \Rightarrow \boxed{\sum_{n=1}^{2020}n^n \equiv_{10} 4}$