Reducing $2018^{2018}\pmod6$ via $\mod2$ and $\mod3$ I've been tasked to reduce $$2018^{2018}\pmod6$$using the results of reducing$$2018^{2018}\pmod2$$and $$2018^{2018}\pmod3$$

I have deduced that $$2018^{2018}\equiv0\pmod2$$and $$2018^{2018}\equiv1\pmod3$$but I don't see a way to combine these two congruences into one $\mod6$ congruence. Could someone tell me how this is achieved? I have a feeling the answer is really simple, but I just don't see it. Is there also a more generalized strategy of solving modular congruences via the divisors of the modulo number($2$ and $3$ are divisors of $6$)?

The way I have currently solved this problem is to first reduce $$2018^{2018}\pmod6$$ to $$2^{2018}\pmod6$$Then we look for patterns:$$\begin{align}2^1&\equiv2\pmod6\\2^2&\equiv4\pmod6\\2^3&\equiv2\pmod6\\2^4&\equiv4\pmod6\\&\ \ .\\&\ \ .\\&\ \ .\\2^n&\equiv\begin{cases}4&\text{if }n\equiv0\pmod2\\2&\text{if }n\equiv1\pmod2\end{cases}\pmod6\end{align}$$for $n\in\Bbb N$. In the case of $2018^{2018}$, $2018\equiv0\pmod2$, so $$2018^{2018}\equiv2^{2018}\equiv4\pmod6$$

The Chinese remainder theorem states that, for these congruence equations:
$$
x\equiv 0 \pmod 2\\
x\equiv1\pmod 3
$$
Then different solutions of $x$ are congruent mod $6$, that is, $x_1 \equiv x_2 \pmod 6$.
You have calculated $2018^{2018} \bmod 2$ and $2018^{2018} \bmod 3$, and know that $x_1 = 2018^{2018}$ is a solution.
By trial and error for all the integers $0,1, \cdots, 6-1$, or by direct construction, $x_2 = 4$ is another solution. So
$$2018^{2018} \equiv 4 \pmod 6$$
