Remainder

For odd integer $n$ , we have $x^n+y^n = (x+y)(x^{n-1} - x^{n-2}y + x^{n-3}y^2 - ... + y^{n-1})$ . Therefore,

$\begin{aligned} 15^{23} + 23^{23} & = (15+23)(15^{22} - 15^{21}23 + 15^{20}23^2 - ... +23^{22}) \\ & = {\color{#3D99F6}38}(15^{22} - 15^{21}23 + 15^{20}23^2 - ... +23^{22}) \end{aligned}$

Since $\color{#3D99F6}38$ is divisible by 19, $15^{23} + 23^{23}$ is divisible by 19, therefore the remainder is $\boxed{0}$ .

${ 15 }^{ 23 }+{ 23 }^{ 23 }\quad \equiv { \quad (-4) }^{ 23 }+{ (4) }^{ 23 }\quad \equiv \quad 0\quad (mod\quad 19)$