Library / /1 of 15

, 2021

Algebra Difficulty 3.7 AMC 10/12 Find the answer United States

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

Pick one

Solutions — 2

Solution 1

Answer (E): Note that $8 = 2^3$ , so
$n = 8^{2022} = (2^3)^{2022} = 2^{6066} = (2^2)^{3033} = 4^{3033}.$
Thus
$\frac{n}{4} = \frac{4^{3033}}{4} = 4^{3032}.$

All the other choices are less than $4^{3032}$ . Choices (A) and (D) have the same base but a lesser exponent. Choice (B) is incorrect because $2^{2022} < 4^{2022} < 4^{3032}$ . Choice (C) is incorrect because
$8^{2018} = (4^{\frac{3}{2}})^{2018} = 4^{3027} < 4^{3032}.$

Solution 2

Solution:
Answer (E): Note that $8 = 2^3$ , so
$n = 8^{2022} = (2^3)^{2022} = 2^{6066} = (2^2)^{3033} = 4^{3033}.$
Thus
$\frac{n}{4} = \frac{4^{3033}}{4} = 4^{3032}.$
All the other choices are less than $4^{3032}$ . Choices (A) and (D) have the same base but a lesser exponent. Choice (B) is incorrect because $2^{2022} < 4^{2022} < 4^{3032}$ . Choice (C) is incorrect because
$8^{2018} = (4^{\frac{3}{2}})^{2018} = 4^{3027} < 4^{3032}.$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.