Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AMC 12P Problems/Problem 1"

(→‎See also)
(→‎Solution 1)
Line 15: Line 15:
  
 
== Solution 1==
 
== Solution 1==
−
If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
+
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 10, 4, and 6 are all even, it is a perfect square. Thus, our answer is <math>\box{\textbf{(C)} 4^4 5^4 6^6},</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|before=First question|num-a=2}}
 
{{AMC12 box|year=2002|ab=P|before=First question|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:42, 30 December 2023

Problem

Which of the following numbers is a perfect square?

$\text{(A) }4^4 5^5 6^6 \qquad \text{(B) }4^4 5^6 6^5 \qquad \text{(C) }4^5 5^4 6^6 \qquad \text{(D) }4^6 5^4 6^5 \qquad \text{(E) }4^6 5^5 6^4$

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 $4^5=2^2^5=2^10$ (Error compiling LaTeX. Unknown error_msg), and since 10, 4, and 6 are all even, it is a perfect square. Thus, our answer is $\box{\textbf{(C)} 4^4 5^4 6^6},$ (Error compiling LaTeX. Unknown error_msg)

See also

2002 AMC 12P (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 12 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12P_Problems/Problem_1&oldid=208897"