<s>[INST] Find $x$ such that $\sum_{k=1}^{2014} k^k \equiv x \pmod {10}$ Find $x$ such that $$\sum_{k=1}^{2014} k^k \equiv x \pmod {10}$$

I knew the answer was $3$.
 [/INST] This is asking for the last digit of that sum. So we may ignore everything but the last digit of each number. Now eliminate the few last terms for a minute and consider:
$$(1^1+\cdots +9^9)+(1^{11}+\cdots +9^{19})+\cdots+(1^{2001}+\cdots +9^{2009})\equiv \,?\pmod{10}$$
Notice that we can now pair each digit and form geometric progressions:
$$\begin{align*}&\quad(1^1+\cdots +1^{2001})+(2^{2}+\cdots +2^{2002})+\cdots+(9^{9}+\cdots +9^{2009})\\[7pt]&=1(1^0+1^{10}+\cdots+1^{2000})+2^2(2^0+2^{10}+\cdots+2^{2000})+9^9(9^0+9^{10}+\cdots 9^{2000})\\[7pt]\end{align*}$$
Rather than writing out the geometric series compactly, let's take it each separately:
$$1^0+1^{10}+\cdots+1^{2000}\equiv 201\equiv 1\pmod{10}$$
$$2^0+2^{10}+\cdots+2^{2000}\equiv 1+\underbrace{4+6+4+\cdots +6}_{200}\equiv 1\pmod{10}$$
$$3^0+3^{10}+\cdots+3^{2000}\equiv 1+\underbrace{9+1+9+\cdots +1}_{200}\equiv 1\pmod{10}$$
$$\cdots$$
$$9^0+9^{10}+\cdots+9^{2000}\equiv 1+\underbrace{1+1+1+\cdots +1}_{200}\equiv 1\pmod{10}$$
Now back to the original expression, we are left with:
$$1+2^2+\cdots+9^9\equiv 7\pmod{10}$$
We have a few more terms to consider, namely:
$$1^{2011}+2^{2012}+3^{2013}+4^{2014}\equiv 6\pmod{10}$$
Hence:
$$\sum_{k=1}^{2014} k^k\equiv 3\pmod{10}$$
(I have left out some of the more trivial computations to avoid clogging this answer with tedious modular arithmetic)
</s>