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Difference between revisions of "2002 AMC 12P Problems/Problem 1"
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<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.
  
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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>.
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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) } 4^5 5^4 6^6}.</math>
  
 
== See also ==
 
== See also ==
 +
{{AMC10 box|year=2002|ab=P|num-b=3|num-a=5}}
 
{{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}}

Latest revision as of 19:37, 28 October 2024

The following problem is from both the 2002 AMC 12P #1 and 2002 AMC 10P #4, so both problems redirect to this page.

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, $\textbf{(C)}$ is a perfect square. Thus, our answer is $\boxed{\textbf{(C) } 4^5 5^4 6^6}.$

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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
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

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