Difference between revisions of "2000 AMC 8 Problems/Problem 2"
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The number <math>0</math> has no reciprocal, and <math>1</math> and <math>-1</math> are their own reciprocals. This leaves only <math>2</math> and <math>-2</math>. The reciprocal of <math>2</math> is <math>1/2</math>, but <math>2</math> is not less than <math>1/2</math>. The reciprocal of <math>-2</math> is <math>-1/2</math>, and <math>-2</math> is less than<math> -1/2</math>, so it is <math>\boxed{A}</math>. | The number <math>0</math> has no reciprocal, and <math>1</math> and <math>-1</math> are their own reciprocals. This leaves only <math>2</math> and <math>-2</math>. The reciprocal of <math>2</math> is <math>1/2</math>, but <math>2</math> is not less than <math>1/2</math>. The reciprocal of <math>-2</math> is <math>-1/2</math>, and <math>-2</math> is less than<math> -1/2</math>, so it is <math>\boxed{A}</math>. | ||
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| + | ==See Also== | ||
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| + | {{AMC8 box|year=2000|before=num-b=1|num-a=3}} | ||
Revision as of 19:05, 14 May 2011
Solution
The number 
has no reciprocal, and 
and 
are their own reciprocals. This leaves only 
and 
. The reciprocal of is 
, but is not less than . The reciprocal of is 
, and is less than, so it is 
.
See Also
| 2000 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by num-b=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 | ||
