Difference between revisions of "2000 AMC 8 Problems/Problem 17"
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+ | ==Problem== | ||
+ | |||
The operation <math> \otimes </math> is defined for all nonzero numbers by <math> a\otimes b =\frac{a^{2}}{b} </math>. Determine <math> [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] </math>. | The operation <math> \otimes </math> is defined for all nonzero numbers by <math> a\otimes b =\frac{a^{2}}{b} </math>. Determine <math> [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] </math>. | ||
<math> \text{(A)}\ -\frac{2}{3}\qquad\text{(B)}\ -\frac{1}{4}\qquad\text{(C)}\ 0\qquad\text{(D)}\ \frac{1}{4}\qquad\text{(E)}\ \frac{2}{3} </math> | <math> \text{(A)}\ -\frac{2}{3}\qquad\text{(B)}\ -\frac{1}{4}\qquad\text{(C)}\ 0\qquad\text{(D)}\ \frac{1}{4}\qquad\text{(E)}\ \frac{2}{3} </math> | ||
+ | |||
+ | ==Solution== | ||
+ | Follow PE(MD)(AS), doing the innermost parentheses first. | ||
+ | |||
+ | <math> [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] </math> | ||
+ | |||
+ | <math> [\frac{1^2}{2}\otimes 3]-[1\otimes \frac{2^2}{3}] </math> | ||
+ | |||
+ | <math> [\frac{1}{2}\otimes 3]-[1\otimes \frac{4}{3}] </math> | ||
+ | |||
+ | <math> [\frac{(\frac{1}{2})^{2}}{3}]-[\frac{1^2}{(\frac{4}{3})}] </math> | ||
+ | |||
+ | <math> [\frac{1}{4} \cdot \frac{1}{3}]-[\frac{3}{4}] </math> | ||
+ | |||
+ | <math>\frac{1}{12} - \frac{3}{4}</math> | ||
+ | |||
+ | <math>\frac{1}{12} - \frac{9}{12}</math> | ||
+ | |||
+ | <math>\frac{-8}{12}</math> | ||
+ | |||
+ | <math>-\frac{2}{3}</math>, which is answer <math>\boxed{A}</math> | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC8 box|year=2000|num-b=16|num-a=18}} |
Revision as of 19:45, 30 July 2011
Problem
The operation is defined for all nonzero numbers by . Determine .
Solution
Follow PE(MD)(AS), doing the innermost parentheses first.
, which is answer
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |