New year's first post

Let $N(n)= \underbrace{20192019\cdots 20192019}_{\text{Number of 2019s }= n}$ . Since $\gcd(N(n), 72) \ne 1$ , we consider $N(n) \bmod 8$ and $N(n) \bmod 9$ separately using Chinese remainder theorem .

For factor 8: $N(n) \equiv {\color{#3D99F6}10^4 N(n-1)} + 2019 \equiv 3 \text{ (mod 8)}$ . Therefore, $N(n) \equiv 8m+3$ , where $m$ is an integer.

For factor 9: We note that the sum of digits of $N(n)$ is $12n$ . Since $12n$ is divisible by 9, when $n$ is a multiple of 3, then $N(n) \bmod 9 = 0$ , when $3\mid n$ . Then there are three cases:

$\begin{cases} \text{If } n \bmod 3 = 0 & \implies N(n) \equiv 12 n \equiv 0 \text{ (mod 9)} & \implies 8m+3 \equiv 0 \text{ (mod 9)} \implies m = 3 \implies \phi (3) \bmod 72 = 27 \\ \text{If } n \bmod 3 = 1 & \implies N(n) \equiv 12n \equiv 3 \text{ (mod 9)} & \implies 8m+3 \equiv 3 \text{ (mod 9)} \implies m = 0 \implies \phi (1) \bmod 72 = 3 \\ \text{If } n \bmod 3 = 2 & \implies N(n) \equiv 12n \equiv 6 \text{ (mod 9)} & \implies 8m+3 \equiv 6 \text{ (mod 9)} \implies m = 6 \implies \phi (2) \bmod 72 = 51 \end{cases}$

Therefore,

$\begin{aligned} x & = -\phi^2(2017) + \phi^2 (2018) - \phi^2(2019) + 2(\phi(2019) + \phi(2018)) \\ & = -\phi^2(1) + \phi^2 (2) - \phi^2(3) + 2(\phi(3) + \phi(2)) \\ & = -3^2 + 51^2 - 27^2 + 2(27 + 51) \\ & = \boxed{2019} \end{aligned}$