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$\begin{aligned} 30^{72^{87}} & \equiv 30^{72^{87} \color{#3D99F6}{\text{ mod 10}}} \text{ (mod 11)} \quad \quad \small \color{#3D99F6}{\text{By Euler's Theorem, since } \gcd(30,11)=1; \ \phi(11) = 10} \\ & \equiv 30^{\color{#3D99F6}{2^{87} \text{ mod 10}}} \text{ (mod 11)} \quad \quad \small \color{#3D99F6}{2^n \text{ mod }10 = \begin{cases} 2^{n \text{ mod }4 } & \text{if }4 \not{|} \ n \\ 6 & \text{if } 4 \ | \ n \end{cases} - \text{ it is cyclical}} \\ & \equiv 30^{\color{#3D99F6}{2^3}} \text{ (mod 11)} \\ & \equiv 30^8 \text{ (mod 11)} \\ & \equiv (33-3)^8 \text{ (mod 11)} \\ & \equiv (-3)^8 \equiv 3^8 \equiv \left(3^2\right)^4 \equiv \left(-2\right)^4 \equiv 16 \equiv \boxed{5} \text{ (mod 11)} \end{aligned}$

Disclaimer: My solution involves Bashing.

Observe that, since 30 and 11 are coprime, we can use Euler's theorem.

$\therefore 30^{\phi (11)} \equiv 1 (\mod 11)$

$\Rightarrow 30^{10} \equiv 1 (\mod 11)$

$\therefore 30^{72^{87}} \equiv 30^{72^{87} (\mod 10)} (\mod 11)$

Applying Euler's theorem again,

$72^{87} \equiv 72^7 \equiv 2^7 \equiv 8 (\mod 10)$

$\Rightarrow 30^{72^{87}} \equiv 30^8 (\mod 11)$

$30 \equiv 1 (\mod 11)$

$30^2 \equiv 9 (\mod 11)$

$30^3 \equiv 6 (\mod 11)$

$30^4 \equiv 4 (\mod 11)$

$30^5 \equiv 10 (\mod 11)$

$30^6 \equiv 3 (\mod 11)$

$30^7 \equiv 2 (\mod 11)$

$30^8 \equiv \boxed{5} (\mod 11)$

$\therefore 30^{72^{87}} \equiv 5(\mod 11)$

And we're done.

Google has several different solutions for this same problem.