remainder

we know a^n + b^n is exactly divisible by a+b when n is odd here n is odd therfore the whole no. is divisible by 23+15 =38 38 is a multiple of 19 hence it is also divisible by 19 rem.=0

When we apply binomial theorem, odd terms get cancelled and even terms remain. it is perfectly divisible by 19

Let the expression given be $Z$ . So, $Z=15^{23}+23^{23}$ . The expression can be rewritten as,

$Z=(19-4)^{23}+(19+4)^{23} \\ \implies Z\equiv (0-4)^{23}+(0+4)^{23} \pmod{19} \\ \implies Z\equiv -4^{23}+4^{23} \pmod{19} \\ \implies Z\equiv 0 \pmod{19}$

Hence, the remainder when the given expression is divided by $19$ is $\boxed{0}$