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 17"

(Added solution and see-also)
(→‎Video Solution)
 
(3 intermediate revisions by 2 users not shown)
Line 25: Line 25:
  
 
<math>-\frac{2}{3}</math>, which is answer <math>\boxed{A}</math>
 
<math>-\frac{2}{3}</math>, which is answer <math>\boxed{A}</math>
 +
 +
 +
== Video Solution ==
 +
 +
https://www.youtube.com/watch?v=dJmKnoKqjX4    ~David
  
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2000|num-b=16|num-a=18}}
 
{{AMC8 box|year=2000|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Latest revision as of 19:05, 15 April 2023

Problem

The operation $\otimes$ is defined for all nonzero numbers by $a\otimes b =\frac{a^{2}}{b}$. Determine $[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)]$.

$\text{(A)}\ -\frac{2}{3}\qquad\text{(B)}\ -\frac{1}{4}\qquad\text{(C)}\ 0\qquad\text{(D)}\ \frac{1}{4}\qquad\text{(E)}\ \frac{2}{3}$

Solution

Follow PE(MD)(AS), doing the innermost parentheses first.

$[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)]$

$[\frac{1^2}{2}\otimes 3]-[1\otimes \frac{2^2}{3}]$

$[\frac{1}{2}\otimes 3]-[1\otimes \frac{4}{3}]$

$[\frac{(\frac{1}{2})^{2}}{3}]-[\frac{1^2}{(\frac{4}{3})}]$

$[\frac{1}{4} \cdot \frac{1}{3}]-[\frac{3}{4}]$

$\frac{1}{12} - \frac{3}{4}$

$\frac{1}{12} - \frac{9}{12}$

$\frac{-8}{12}$

$-\frac{2}{3}$, which is answer $\boxed{A}$


Video Solution

https://www.youtube.com/watch?v=dJmKnoKqjX4 ~David

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_17&oldid=192209"