It is 2018, is it not?

Define $Q(x)$ such that $Q(x)=P(x)-2018$ .

Thus,

• $Q(1)=0$

• $Q(2)=0$

• $Q(3)=0$

• $Q(4)=-36$

Because $Q(1)=0$ , then $(x-1)$ is a factor of the polynomial $Q(x)$ . The same logic holds for $Q(2)$ and $Q(3)$ .

Since $Q(x)$ is a three degree polynomial, (as we know from $P(x)$ ), then $Q(x)=h(x-1)(x-2)(x-3)$ , where h is some constant.

Plugging $x=4$ in, we get $h=-6 \longrightarrow Q(x)=-6(x-1)(x-2)(x-3)$

Therefore, $P(x)=Q(x)+2018=-6(x-1)(x-2)(x-3)+2018$ .

Plugging in $x=0$ , we get an answer of $36+2018=\boxed{2054}$

We note that $P(x) = k(x-1)(x-2)(x-3) + 2018$ satisfies $P(1)=P(2)=P(3) = 2018$ . If $P(4)=1982$ , then:

$\begin{aligned} k(4-1)(4-2)(4-3)+2018 & = 1982 \\ 6 k & = 1982 - 2018 \\ \implies k & = - 6 \\ \implies P(x) & = 2018 - 6(x-1)(x-2)(x-3) \\ P(0) & = 2018 -6(-6) = \boxed{2054} \end{aligned}$