2022 AMC 10B Problems/Problem 1: Difference between revisions
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{{duplicate|[[2022 AMC 10B Problems/Problem 1|2022 AMC 10B #1]] and [[2022 AMC 12B Problems/Problem 1|2022 AMC 12B #1]]}} | |||
==Problem== | |||
Define <math>x\diamondsuit y</math> to be <math>|x-y|</math> for all real numbers <math>x</math> and <math>y.</math> What is the value of <cmath>(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)?</cmath> | |||
<math>\textbf{(A)}\ {-}2 \qquad\textbf{(B)}\ {-}1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 2</math> | |||
== Solution == | |||
We have <cmath>(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)=(1\diamondsuit1)-(1\diamondsuit3)=0-2=\boxed{\textbf{(A)}\ {-}2}.</cmath> | |||
~MRENTHUSIASM | |||
== See Also == | |||
{{AMC10 box|year=2022|ab=B|before=First Problem|num-a=2}} | |||
{{AMC12 box|year=2022|ab=B|before=First Problem|num-a=2}} | |||
{{MAA Notice}} | |||
Revision as of 14:24, 17 November 2022
- The following problem is from both the 2022 AMC 10B #1 and 2022 AMC 12B #1, so both problems redirect to this page.
Problem
Define 
to be 
for all real numbers 
and 
What is the value of ![\[(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)?\]](http://latex.artofproblemsolving.com/e/4/8/e486594a9c3be11fd5275aa45b53a4aecf616093.png)
Solution
We have ![\[(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)=(1\diamondsuit1)-(1\diamondsuit3)=0-2=\boxed{\textbf{(A)}\ {-}2}.\]](http://latex.artofproblemsolving.com/5/7/1/57150d8c9cde5436d44e6843b310bdcb780e2958.png)
~MRENTHUSIASM
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
