Difference between revisions of "2022 AMC 10B Problems/Problem 1"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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+ | == Solution 2 == | ||
+ | Observe that the <math>\diamond</math> function is simply the positive difference between two numbers. Thus, we evaluate: the difference between <math>2</math> and <math>3</math> is <math>1</math>; the difference between <math>1</math> and <math>1</math> is <math>0</math>; the difference between <math>1</math> and <math>2</math> is <math>1</math>; the difference between <math>1</math> and <math>3</math> is <math>2</math>; and finally, <math>0-2=\boxed{\text{(A)}~-2.</math> | ||
+ | |||
+ | ~Technodoggo | ||
==Video Solution (⚡️Solved in 50 seconds⚡️)== | ==Video Solution (⚡️Solved in 50 seconds⚡️)== |
Revision as of 01:30, 23 October 2023
- The following problem is from both the 2022 AMC 10B #1 and 2022 AMC 12B #1, so both problems redirect to this page.
Contents
Problem
Define to be for all real numbers and What is the value of
Solution
We have ~MRENTHUSIASM
Solution 2
Observe that the function is simply the positive difference between two numbers. Thus, we evaluate: the difference between and is ; the difference between and is ; the difference between and is ; the difference between and is ; and finally, $0-2=\boxed{\text{(A)}~-2.$ (Error compiling LaTeX. Unknown error_msg)
~Technodoggo
Video Solution (⚡️Solved in 50 seconds⚡️)
~Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
Video Solution by Interstigation
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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 |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by First Problem |
Followed by Problem 2 |
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.