Remainders

Relevant wiki: Fermat's Little Theorem

We use Fermat's Little Theorem which states that $a^{p-1}\equiv 1\bmod p$ for prime $p$ and $a$ not divisible by $p$ . Equivalently, $a^n \equiv a^{n \bmod (p-1)}\bmod p$ . Thus $\begin{aligned} 2^{5^9} + 5^{9^2} + 9^{2^5} &\equiv 2^{5^9 \bmod 10} + 5^{9^2 \bmod 10} +9^{2^5\bmod 10} && \bmod 11\\ &\equiv 2^5 + 5^1 + 9^2 && \bmod 11 \\ &\equiv -1 + 5 + 4 && \bmod 11 \\ &\equiv \boxed{8} && \bmod 11 \end{aligned}$

Relevant wiki: Euler's Theorem

Similar solution with @Jordan Cahn's

$\begin{aligned} 2^{5^9} + 5^{9^2} + 9^{2^5} & \equiv 2^{\color{#3D99F6}5^9 \bmod \phi(11)} + 5^{\color{#3D99F6}9^2 \bmod \phi(11)} + 9^{\color{#3D99F6}2^5 \bmod \phi(11)} \text{ (mod 11)} & \small \color{#3D99F6} \text{Since }\gcd (2,11) = \gcd (5,11) = \gcd(9,11) = 1 \text{, Euler's theorem applies.} \\ & \equiv 2^{\color{#3D99F6}5^9 \bmod 10} + 5^{\color{#3D99F6}9^2 \bmod 10} + 9^{\color{#3D99F6}2^5 \bmod 10} \text{ (mod 11)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (11) = 11-1 = 10 \\ & \equiv 2^{\color{#3D99F6}5} + 5^{(10-1)^2 \bmod 10} + 9^{32 \bmod 10} \text{ (mod 11)} & \small \color{#3D99F6} \text{Powers of 5 always end with 25 } \implies 5^n \bmod 10 = 5 \\ & \equiv 2^5 + 5^1 + 9^2 \text{ (mod 11)} \\ & \equiv 32 + 5 + 81 \text{ (mod 11)} \\ & \equiv -1 + 5 + 4 \text{ (mod 11)} \\ & \equiv \boxed 8 \text{ (mod 11)} \end{aligned}$