2002 AMC 10P Problems/Problem 1

Problem 1

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

For a positive integer to be a perfect square, all the primes in its prime factorization must have an even exponent. With a quick glance at the answer choices, we can eliminate options

$\textbf{(A)}$ because $5^5$ is an odd power

$\textbf{(B)}$ because $6^5 = 2^5 \cdot 3^5$ and $3^5$ is an odd power

$\textbf{(D)}$ because $6^5 = 2^5 \cdot 3^5$ and $3^5$ is an odd power, and

$\textbf{(E)}$ because $5^5$ is an odd power.

This leaves option $\textbf{(C)},$ in which $4^5=(2^{2})^{5}=2^{10}$ , and since $10, 4,$ and $6$ are all even, $\textbf{(C)}$ is a perfect square. Thus, our answer is $\boxed{\textbf{(C) } 4^4 5^4 6^6}$ .

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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All AMC 10 Problems and Solutions

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2002 AMC 10P Problems/Problem 1

Problem 1

Solution 1

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