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

Revision as of 17:36, 14 July 2024

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-== Problem ==
+== Problem 1 ==
-Which of the following numbers is a perfect square?
+The ratio <math>\frac{(2^4)^8}{(4^8)^2}</math> equals
 <math>
-\text{(A) }4^4 5^5 6^6
+\text{(A) }\frac{1}{4}
 \qquad
-\text{(B) }4^4 5^6 6^5
+\text{(B) }\frac{1}{2}
 \qquad
-\text{(C) }4^5 5^4 6^6
+\text{(C) }1
 \qquad
-\text{(D) }4^6 5^4 6^5
+\text{(D) }2
 \qquad
-\text{(E) }4^6 5^5 6^4
+\text{(E) }8
 </math>

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

Revision as of 17:36, 14 July 2024

Problem 1

Solution 1

See also