Difference between revisions of "2000 AMC 12 Problems/Problem 11"
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==Solution 2== | ==Solution 2== | ||
− | This simplifies to <math>ab+b-a=0 \Rightarrow (a+1)(b-1) = -1</math>. The two integer solutions to this are <math>(-2,2)</math> and <math>(0,0)</math>. The problem states than <math>a</math> and <math>b</math> are non-zero, so we consider the case of <math>(-2,2)</math>. So, we end up with <math>\frac{-2}{2} + \frac{2}{-2} - 2 \cdot -2 = 4 - 1 - 1 = 2 \ | + | This simplifies to <math>ab+b-a=0 \Rightarrow (a+1)(b-1) = -1</math>. The two integer solutions to this are <math>(-2,2)</math> and <math>(0,0)</math>. The problem states than <math>a</math> and <math>b</math> are non-zero, so we consider the case of <math>(-2,2)</math>. So, we end up with <math>\frac{-2}{2} + \frac{2}{-2} - 2 \cdot -2 = 4 - 1 - 1 = 2 \Rightarrow \boxed{\text{E}}</math> |
==Video Solution== | ==Video Solution== |
Latest revision as of 17:21, 19 October 2020
- The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.
Problem
Two non-zero real numbers, and satisfy . Which of the following is a possible value of ?
Solution 1
.
Another way is to solve the equation for giving then substituting this into the expression and simplifying gives the answer of
Solution 2
This simplifies to . The two integer solutions to this are and . The problem states than and are non-zero, so we consider the case of . So, we end up with
Video Solution
https://www.youtube.com/watch?v=8nxvuv5oZ7A&t=3s
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 AMC 12 Problems and Solutions |
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.