The remainder when $(2021)^{2022} + (2022)^{2021}$ is divided by $7$ is:

The numbers $a_n = 6^n - 5n$ for $n = 1, 2, 3, \ldots$ when divided by $25$ leave the remainder:

For each $n \in N$, $2^{3n}-1$ is divisible by, where $N$ is the set of natural numbers.

The remainder when $(2023)^{2023}$ is divided by $35$ is $..........$.

The remainder when $7^{7^{7^{...7}}}$ ($22$ times $7$) is divided by $48$ is

Let the number $(22)^{2022} + (2022)^{22}$ leave the remainder $\alpha$ when divided by $3$ and $\beta$ when divided by $7$. Then $(\alpha^2 + \beta^2)$ is equal to