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Difference between revisions of "2022 AMC 10B Problems/Problem 1"

m (→‎Problem)
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==Problem==
 
==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>
+
Define <math>x\diamond 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\diamond(2\diamond3))-((1\diamond2)\diamond3)?</cmath>
  
 
<math> \textbf{(A)}\ {-}2 \qquad
 
<math> \textbf{(A)}\ {-}2 \qquad
Line 12: Line 12:
 
== Solution ==  
 
== Solution ==  
 
We have <cmath>\begin{align*}
 
We have <cmath>\begin{align*}
−
(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3) &= |1-|2-3|| - ||1-2|-3| \\
+
(1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\
 
&= |1-1| - |1-3| \\
 
&= |1-1| - |1-3| \\
 
&= 0-2 \\
 
&= 0-2 \\

Revision as of 15:56, 6 March 2023

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 $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]

$\textbf{(A)}\ {-}2 \qquad \textbf{(B)}\ {-}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$

Solution

We have \begin{align*} (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ &= |1-1| - |1-3| \\ &= 0-2 \\ &=\boxed{\textbf{(A)}\ {-}2}. \end{align*} ~MRENTHUSIASM

Video Solution 1

https://youtu.be/4xOeX0aQF3U

~Education, the Study of Everything

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. AMC logo.png

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