Sum of Towers

Relevant wiki: Euler's Theorem

Let the given number be $N$ .

$\large \begin{aligned} N & \equiv 10^{5^{10^{5^{10}}}} + 5^{\color{#3D99F6}10^{5^{10^5}} \bmod \phi(11)} \text{ (mod 11)} & \small \color{#3D99F6} \text{Since }\gcd(5,11) = 1 \text{, Euler's theorem applies.} \\ & \equiv (11-1)^{\color{#D61F06}5^{10^{5^{10}}}} + 5^{\color{#3D99F6}10^{5^{10^5}} \bmod 10} \text{ (mod 11)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(11) = 10 \\ & \equiv {\color{#D61F06}-1} + + 5^{\color{#3D99F6}0} \text{ (mod 11)} & \small \color{#D61F06} \text{Note that }5^n \text{ is odd } \forall \ n \in \mathbb N \\ & \equiv -1 + 1 \text{ (mod 11)} \\ & \equiv \boxed{0} \text{ (mod 11)} \end{aligned}$

Yes , the number $N$ is divisible by 11.

$10^{5^{10^{5^{10}}}} \mod{11} \equiv (-1)^{5^{10^{5^{10}}}} \mod{11} \equiv -1 \mod{11}$

Fermat's little theorem tells us that $5^{10} \equiv 1 \mod{11}$ , so

$5^{10^{5^{10^{5}}}} \equiv 1 \mod{11} \Rightarrow 10^{5^{10^{5^{10}}}}+5^{10^{5^{10^{5}}}} \mod{11} \equiv (-1)+(1) \mod{11} \equiv 0 \mod{11}$