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 $$2018^{2018}\equiv2^{2018}\equiv4\pmod6$$

Let $x=2018^{2018},$ then
$$ \qquad\qquad x\equiv (-1)^{2018}\equiv 1 \quad  \pmod 3 \tag*{(1)}$$
$$ x\equiv 0 \quad  \pmod 2 \tag*{(2)} $$
From $(1)$, we have $(*):x=3m+1$ for some integer $m$. Then putting $x$ into $(2)$ yields $$ \begin{aligned} & \quad  3m+1\equiv 0 \quad  \pmod 2\\\Leftrightarrow & \quad  m\equiv 1 \quad  \pmod 2 \\\Leftrightarrow &  \quad  m=2n+1 \textrm{ for some integer }n \end{aligned}$$
Putting back to (*) conclude that $$x=3(2n+1)+1\equiv 4 \quad  \pmod 6.$$
