F-Actor?

Let the given number be $N$ . We need to find if $N \equiv 0 \text{ (mod 8)}$ .

$\begin{aligned} N & \equiv 23^{2017} + 17^{2018} \text{ (mod 8)} \\ & \equiv (24-1)^{2017} + (16+1)^{2018} \text{ (mod 8)} \\ & \equiv (-1)^{2017} + 1^{2018} \text{ (mod 8)} \\ & \equiv -1 + 1 \text{ (mod 8)} \\ & \equiv \boxed 0 \text{ (mod 8)} \end{aligned}$

Yes , 8 is a factor of $N$ .

Take the polynomial $P(x)=(3x-7)^{2017}+(2x-3)^{2018}$ it obviously has $(x-2)$ as a factor because $P(2)=0$ We have $P(x) \equiv 0 \mod (x-2)$ By substituting $x=10$ we get $P(10) \equiv 0 \mod 8$ or $23^{2017}+17^{2018} \equiv 0 \mod 8$ Hence, 8 is a factor