Difference between revisions of "2002 AMC 10P Problems/Problem 1"

Latest revision as of 17:47, 18 October 2025

The ratio $\frac{(2^4)^8}{(4^8)^2}$ equals

$\text{(A) }\frac{1}{4} \qquad \text{(B) }\frac{1}{2} \qquad \text{(C) }1 \qquad \text{(D) }2 \qquad \text{(E) }8$

We can use basic rules of exponentiation to solve this problem.

$\frac{(2^4)^8}{(4^8)^2} =\frac{(2^4)^8}{(2^{16})^2} =\frac{2^{32}}{2^{32}} =1$

Thus, our answer is $\boxed{\textbf{(C) } 1}.$

We can rearrange the exponents on the bottom to solve this problem:

$\frac{(2^4)^8}{(4^8)^2} =\frac{(2^4)^8}{(4^{2})^8} =\frac{16^{8}}{16^{8}} =\boxed{\textbf{(C) } 1}$

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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 {{AMC10 box|year=2002|ab=P|before=First question|num-a=2}}
 {{MAA Notice}}
+[[Category: Introductory Algebra Problems]]