Teens These Days

We seek to find $x$ such that $13^{19} + 17^{19} \equiv x \pmod {15}.$

Fortunately, by rules of modular arithmetic , we know that

$13^{19} \equiv (-2)^{19} \pmod{15}.$

Similarly, $17^{19} \equiv 2^{19} \pmod{15}.$

Therefore,

$13^{19} + 17^{19} \equiv (-2)^{19} + 2^{19} \equiv \boxed{0} \pmod {15}.$

a^n + b^n is divisible by a+b, provided n is odd. Here it is 30. So obviously, remainder is 0.

$13^{19} + 17^{19} = (13+17)(13^{18} - 13^{17}.17 + \dots -17^{17}.13 +17^{18}) = 30\times (sum) \rightarrow \text{multiple of 15}$

I used $a^n+b^n = (a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - \dots +a^2b^{n-3} - ab^{n-2}+b^{n-1})$ for odd $n>1$ .