2021 AMC Fall 10B Problem 5

The numbers $8$ and $4$ are both powers of $2$, so rewrite everything with base $2$.

1. $8 = 2^3$, so
$$n = 8^{2022} = (2^3)^{2022} = 2^{3 \times 2022} = 2^{6066}.$$
2. $4 = 2^2$. Dividing by $2^2$ takes $2$ off the exponent:
$$\frac{n}{4} = \frac{2^{6066}}{2^2} = 2^{6064}.$$
3. The choices are mostly powers of $4$. Since $4 = 2^2$, every two factors of $2$ make one factor of $4$. So $6064$ twos make $6064 \div 2 = 3032$ fours:
$$2^{6064} = (2^2)^{3032} = 4^{3032}.$$

The answer is $\boxed{\textbf{(E)}\ 4^{3032}}$.

When powers have different bases that are all made from the same small number, change everything to that small number first. Then multiplying and dividing become adding and subtracting exponents. At the end, switch to whatever base the answer choices use.

Express everything in base $2$. Since $8 = 2^3$,
$$n = 8^{2022} = (2^3)^{2022} = 2^{6066}.$$
Dividing by $4 = 2^2$ subtracts $2$ from the exponent:
$$\frac{n}{4} = 2^{6066 - 2} = 2^{6064}.$$
The choices are powers of $4$, and $2^{6064} = (2^2)^{3032} = 4^{3032}$.

The answer is $\boxed{\textbf{(E)}\ 4^{3032}}$.