Difference between revisions of "2002 AMC 12P Problems/Problem 1"
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+ | {{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #1]] and [[2002 AMC 10P Problems|2002 AMC 10P #4]]}} | ||
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== Problem == | == Problem == | ||
Which of the following numbers is a perfect square? | Which of the following numbers is a perfect square? | ||
Line 25: | Line 27: | ||
<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, 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^ | + | 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?
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, is a perfect square. Thus, our answer is
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.