Difference between revisions of "2022 AMC 10B Problems/Problem 1"
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\textbf{(E)}\ 2</math> | \textbf{(E)}\ 2</math> | ||
| − | == Solution == | + | == Solution 1 == |
We have <cmath>\begin{align*} | We have <cmath>\begin{align*} | ||
(1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ | (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ | ||
Latest revision as of 03:22, 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 ![\[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]](http://latex.artofproblemsolving.com/3/0/a/30abfad2fabbf18f3fae4d47e45fdcc4bf8a8d94.png)
Solution 1
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, 
~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. 
