2002 AMC 10P Problems/Problem 1

Revision as of 18:15, 14 July 2024 by Wes (talk | contribs) (Solution 1)

Problem

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

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

Solution 1

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

\begin{align*} \frac{(2^4)^8}{(4^8)^2}$\\ =\frac{(2^4)^8}{(2^16)^2}$\\ =\frac{2^32}{2^32}$\\ =1 \end{align*}

Thus, our answer is$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(C) } 1}.$

See also

2002 AMC 10P (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
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All AMC 10 Problems and Solutions

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