2002 AMC 12P Problems/Problem 1: Difference between revisions
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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. | ||
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,</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>. | ||
== 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?
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
because 
is an odd power
because 
and 
is an odd power
because and is an odd power, and
because is an odd power.
This leaves option 
in which 
, and since 
and 
are all even, it is a perfect square. Thus, our answer is 
.
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
