Difference between revisions of "2000 AMC 12 Problems/Problem 5"
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Thus <math>x-p = (2-p)-p = 2-2p</math>. | Thus <math>x-p = (2-p)-p = 2-2p</math>. | ||
<math>\boxed{\mathbf{(C)}\ \ensuremath{2-2p}}</math> | <math>\boxed{\mathbf{(C)}\ \ensuremath{2-2p}}</math> | ||
| + | == Solution 2 (guess and check/desperation) == | ||
| + | |||
| + | If you did not find that slick Solution 1, all hope is not lost. We could still guess and check our way to the right answer. | ||
| + | |||
| + | We first plug in <math>x=1</math>, and get that <math>p=1</math> too. Hence <math>x-p=0</math>, eliminating choices <math>A</math> and <math>B</math>. | ||
| + | |||
| + | We then plug in <math>x=0</math>, and get <math>p=2</math>. Therefore, <math>x-p=-2</math>. The answer is negative, eliminating <math>E</math>. Furthermore, <math>2p-2=2(2)-2=4-2=2\neq-2</math>, so choice <math>D</math> is false. Hence, the answer must be <math>C</math>, which upon checking indeed still holds true. | ||
| + | -Monkey_King | ||
==Video Solution by Daily Dose of Math== | ==Video Solution by Daily Dose of Math== | ||
Latest revision as of 23:20, 5 September 2024
- The following problem is from both the 2000 AMC 12 #5 and 2000 AMC 10 #9, so both problems redirect to this page.
Contents
Problem
If 
, where 
, then 
Solution
When 
is negative so 
and 
.
Thus 
.
Solution 2 (guess and check/desperation)
If you did not find that slick Solution 1, all hope is not lost. We could still guess and check our way to the right answer.
We first plug in 
, and get that 
too. Hence 
, eliminating choices 
and 
.
We then plug in 
, and get 
. Therefore, 
. The answer is negative, eliminating 
. Furthermore, 
, so choice 
is false. Hence, the answer must be 
, which upon checking indeed still holds true.
-Monkey_King
Video Solution by Daily Dose of Math
https://youtu.be/albUhCOwv3Y?si=4XcusOEp70EA6XKr
~Thesmartgreekmathdude
See also
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
