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

Revision as of 17:36, 14 July 2024

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

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

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

@@ Line 15: / Line 15: @@
 == 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
-<math>\textbf{(A)}</math> because <math>5^5</math> is an odd power
-<math>\textbf{(B)}</math> because <math>6^5 = 2^5 \cdot 3^5</math> and <math>3^5</math> is an odd power
-<math>\textbf{(D)}</math> because <math>6^5 = 2^5 \cdot 3^5</math> and <math>3^5</math> is an odd power, and
-<math>\textbf{(E)}</math> because <math>5^5</math> is an odd power.
-This leaves option <math>\textbf{(C)},</math> in which <math>4^5=(2^{2})^{5}=2^{10}</math>, and since <math>10, 4,</math> and <math>6</math> are all even, <math>\textbf{(C)}</math> is a perfect square. Thus, our answer is <math>\boxed{\textbf{(C) } 4^4 5^4 6^6}</math>.
 == See also ==
 {{AMC10 box|year=2002|ab=P|before=First question|num-a=2}}
 {{MAA Notice}}