Difference between revisions of "2002 AMC 12P Problems/Problem 1"
(→See also) |
(→Solution 1) |
||
(9 intermediate revisions by the same user not shown) | |||
Line 15: | Line 15: | ||
== Solution 1== | == 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 | |
+ | |||
+ | <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 <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>. | ||
== See also == | == See also == | ||
− | {{AMC12 box|year=2002|ab=P| | + | {{AMC12 box|year=2002|ab=P|before=First question|num-a=2}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:54, 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, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.