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

(Created page with "== Problem == Which of the following numbers is a perfect square? <math> \text{(A) }4^4 5^5 6^6 \qquad \text{(B) }4^4 5^6 6^5 \qquad \text{(C) }4^5 5^4 6^6 \qquad \text{(D) }...")
 
(→‎See also)
 
(19 intermediate revisions by 2 users not shown)
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== Problem ==
 
== Problem ==
−
Which of the following numbers is a perfect square?
+
The ratio <math>\frac{(2^4)^8}{(4^8)^2}</math> equals
  
 
<math>
 
<math>
−
\text{(A) }4^4 5^5 6^6
+
\text{(A) }\frac{1}{4}
 
\qquad
 
\qquad
−
\text{(B) }4^4 5^6 6^5
+
\text{(B) }\frac{1}{2}
 
\qquad
 
\qquad
−
\text{(C) }4^5 5^4 6^6
+
\text{(C) }1
 
\qquad
 
\qquad
−
\text{(D) }4^6 5^4 6^5
+
\text{(D) }2
 
\qquad
 
\qquad
−
\text{(E) }4^6 5^5 6^4
+
\text{(E) }8
 
</math>
 
</math>
  
 
== Solution 1==
 
== 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
+
We can use basic rules of exponentiation to solve this problem.
  
−
<math>\textbf{(A)}</math> because <math>5^5</math> is an odd power
+
<math>\frac{(2^4)^8}{(4^8)^2}
 +
=\frac{(2^4)^8}{(2^{16})^2}
 +
=\frac{2^{32}}{2^{32}}
 +
=1</math>
  
−
<math>\textbf{(B)}</math> because <math>6^5 = 2^5 \cdot 3^5</math> and <math>3^5</math> is an odd power
+
Thus, our answer is <math>\boxed{\textbf{(C) } 1}.</math>
−
 
+
== Solution 2==
−
<math>\textbf{(D)}</math> because <math>6^5 = 2^5 \cdot 3^5</math> and <math>3^5</math> is an odd power, and
+
We can rearrange the exponents on the bottom to solve this problem:
−
 
 
−
<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>.
 
  
 +
<math>\frac{(2^4)^8}{(4^8)^2}
 +
=\frac{(2^4)^8}{(4^{2})^8}
 +
=\frac{16^{8}}{16^{8}}
 +
=\boxed{\textbf{(C) } 1}</math>
 
== See also ==
 
== See also ==
−
{{AMC12 box|year=2002|ab=P|before=First question|num-a=2}}
+
{{AMC10 box|year=2002|ab=P|before=First question|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category: Introductory Algebra Problems]]

Latest revision as of 16:47, 18 October 2025

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.

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

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

Solution 2

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

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

See also

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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