Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AMC 8 Problems/Problem 2"
Line 5: Line 5:
 
==See Also==
 
==See Also==
  
−
{{AMC8 box|year=2000|before=num-b=1|num-a=3}}
+
{{AMC8 box|year=2000|num-b=1|num-a=3}}

Revision as of 18:06, 14 May 2011

Solution

The number $0$ has no reciprocal, and $1$ and $-1$ are their own reciprocals. This leaves only $2$ and $-2$. The reciprocal of $2$ is $1/2$, but $2$ is not less than $1/2$. The reciprocal of $-2$ is $-1/2$, and $-2$ is less than$-1/2$, so it is $\boxed{A}$.

See Also

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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 AJHSME/AMC 8 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_2&oldid=38566"