2021 AMC Fall 10B Problem 5

P1-10~1 minPrint

Exponents, Logarithms & Radicals Set up the equation/formula and compute; no special trick needed

Let $n=8^{2022}$ . Which of the following is equal to $\frac{n}{4}?$

Show solution without answering

Key insight

Switch to base 2: 8 to the 2022 is 2 to the 6066; dividing by 4 gives 2 to the 6064, which is 4 to the 3032.

Solution

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.4)
Introduction to Algebra, Chapter 19: Exponents and Logarithms (§19.4)