Difference between revisions of "2001 AMC 8 Problems/Problem 12"
(→Solution) |
|||
(One intermediate revision by one other user not shown) | |||
Line 9: | Line 9: | ||
<math> 6\otimes4=\frac{6+4}{6-4}=5 </math>. | <math> 6\otimes4=\frac{6+4}{6-4}=5 </math>. | ||
<math> 5\otimes3=\frac{5+3}{5-3}=4, \boxed{\text{A}} </math> | <math> 5\otimes3=\frac{5+3}{5-3}=4, \boxed{\text{A}} </math> | ||
+ | |||
+ | ==Video Solution-Cooler Method== | ||
+ | https://www.youtube.com/watch?v=ZfwtAiH_6PI&t=36s | ||
==See Also== | ==See Also== | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
{{AMC8 box|year=2001|num-b=11|num-a=13}} | {{AMC8 box|year=2001|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:42, 26 December 2023
Problem
If , then
Solution
.
Video Solution-Cooler Method
https://www.youtube.com/watch?v=ZfwtAiH_6PI&t=36s
See Also
2001 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.