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

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==Problem==
 
==Problem==
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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>
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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>
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<math>\textbf{(A)}\ {-}2 \qquad\textbf{(B)}\ {-}1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 2</math>
 
  
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== Solution ==  
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<math> \textbf{(A)}\ {-}2 \qquad
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We have <cmath>(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)=(1\diamondsuit1)-(1\diamondsuit3)=0-2=\boxed{\textbf{(A)}\ {-}2}.</cmath>
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\textbf{(B)}\ {-}1 \qquad
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\textbf{(C)}\ 0 \qquad
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\textbf{(D)}\ 1 \qquad
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\textbf{(E)}\ 2</math>
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== Solution 1 ==  
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We have <cmath>\begin{align*}
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(1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\
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&= |1-1| - |1-3| \\
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&= 0-2 \\
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&= \boxed{\textbf{(A)}\ {-}2}.
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\end{align*}</cmath>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
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== Solution 2 (Substitution) ==
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== Solution 2 ==
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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{\textbf{(A)}\ {-}2}.</math>  
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<math>(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)</math>  
 
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<math>(|1-|2-3||)-(|1-2|-3|)</math>
 
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<math>(|1-|-1||)-(|-1|-3|)</math>
 
  
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<math>(|1-1|)-(|1-3|)</math>
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~Technodoggo
  
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<math>(|0|)-(|-2|)</math>
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==Video Solution (⚡️Solved in 50 seconds⚡️)==
 +
https://youtu.be/4xOeX0aQF3U
  
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<math>(0)-(2)</math>
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~Education, the Study of Everything
  
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<math>0-2=\boxed{\textbf{(A)}\ {-}2}.</math>
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==Video Solution(1-16)==
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https://youtu.be/SCwQ9jUfr0g
  
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~ghfhgvghj10
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~~Hayabusa1
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==Video Solution by Interstigation==
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https://youtu.be/_KNR0JV5rdI
  
 
== See Also ==
 
== See Also ==

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.

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 1

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

Solution 2

Observe that the $\diamond$ function is simply the positive difference between two numbers. Thus, we evaluate: the difference between $2$ and $3$ is $1;$ the difference between $1$ and $1$ is $0;$ the difference between $1$ and $2$ is $1;$ the difference between $1$ and $3$ is $2;$ and finally, $0-2=\boxed{\textbf{(A)}\ {-}2}.$

~Technodoggo

Video Solution (⚡️Solved in 50 seconds⚡️)

https://youtu.be/4xOeX0aQF3U

~Education, the Study of Everything

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

Video Solution by Interstigation

https://youtu.be/_KNR0JV5rdI

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

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