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

(Solution 1)
(Solution 1)
Line 25: Line 25:
 
<math>\textbf{(E)}</math> because <math>5^5</math> is an odd power.
 
<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,</math> <math>4,</math> and <math>6</math> are all even, it is a perfect square. Thus, our answer is <math>\boxed{\textbf{(C) } 4^4 5^4 6^6}</math>.
+
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, it is a perfect square. Thus, our answer is <math>\boxed{\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 00:53, 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 $4^5=(2^{2})^{5}=2^{10}$, and since $10, 4,$ and $6$ are all even, it is a perfect square. Thus, our answer is $\boxed{\textbf{(C) } 4^4 5^4 6^6}$.

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
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