A number theory problem by Aly Ahmed

Let the given number be $N$ . We need to find $N \bmod 11$ . Since $\gcd(2020,11)=1$ , we can apply Euler's theorem , and Euler's totient function $\phi(11)=11-1=10$ .

$\begin{aligned} N & \equiv 2020^{2021^{2022}\bmod \phi(11)} \text{ (mod 11)} \\ & \equiv 2020^{2021^{2022}\bmod 10} \text{ (mod 11)} \\ & \equiv 2020^{(2020+1)^{2022}\bmod 10} \text{ (mod 11)} \\ & \equiv 2020^{1^{2022}\bmod 10} \text{ (mod 11)} \\ & \equiv 2020^1 \text{ (mod 11)} \\ & \equiv \boxed 7 \text{ (mod 11)} \end{aligned}$

$2020\equiv 7$ (mod $11$ ). So the given congruence is equivalent to

$7^{2021^{2022}}$ (mod $11$ ).

Now, for any positive integers $n, p; 7^n\equiv 7^{10p+n}$ (mod $11$ ).

$2021^{2022}\equiv 1$ (mod $10$ )

Therefore $2020^{2021^{2022}}\equiv 7^1=\boxed 7$ (mod $11$ ).