Remainder?

$\begin{aligned} x & \equiv 2^{100} + 3^{100} + 4^{100} + 5^{100} \text{ (mod 7)} \\ & \equiv 2^{100} + (7-4)^{100} + 4^{100} + (7-2)^{100} \text{ (mod 7)} \\ & \equiv 2^{100} + (-4)^{100} + 4^{100} + (-2)^{100} \text{ (mod 7)} \\ & \equiv 2\times 2^{100} + 2 \times 4^{100} \text{ (mod 7)} \\ & \equiv 2^{101} + 2^{201} \text{ (mod 7)} \\ & \equiv 2^2 \times 2^{3\times 33} + 2^{3\times 67} \text{ (mod 7)} \\ & \equiv 4(7+1)^{33} + (7+1)^{67} \text{ (mod 7)} \\ & \equiv 4 + 1 \equiv \boxed 5 \text{ (mod 7)} \end{aligned}$

By Fermat's Little Theorem , for $a$ not divisible by a prime $p$ , $a^n \equiv a^{n\bmod p-1}\mod p$ . Thus $\begin{aligned} 2^{100}+3^{100}+4^{100} + 5^{100} &\equiv 2^{100 \bmod 6} + 3^{100\bmod 6} + 4^{100\bmod 6} + 5^{100\bmod 6} &\mod 7 \\ &\equiv 2^4 + 3^4 + 4^4 + 5^4 &\mod 7 \\ &\equiv 2 + 4 + 4 + 2 &\mod 7 \\ &\equiv \boxed{5} &\mod 7 \end{aligned}$