How to find the remainder?

By Fermat’s Little Theorem $a^{n-1} \equiv 1\pmod n$ where $n$ is prime. Since 103 is prime, each number in the series is equal to $1\pmod{103}$ so the sum is just $100 \pmod{103}$ .

Since $103$ is a prime, we can apply the Euler's theorem for number $1, 2, 3, \cdots 100$ . The Euler's totient function $\phi(103) = 103 - 1 = 102$ .

$\sum_{n=1}^{100} n^{102} \equiv \sum_{n=1}^{100} n^{102 \bmod \phi(103)} \equiv \sum_{n=1}^{100} n^{102 \bmod 102} \equiv \sum_{n=1}^{100} n^0 \equiv \sum_{n=1}^{100} 1 \equiv \boxed{100} \text{ (mod 103)}$