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Difference between revisions of "2022 AMC 10B Problems/Problem 1"
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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>
 
We have <cmath>(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)=(1\diamondsuit1)-(1\diamondsuit3)=0-2=\boxed{\textbf{(A)}\ {-}2}.</cmath>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
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 +
== Solution 2 (Substiution) ==
 +
 +
<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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<math>(|0|)-(|-2|)</math>
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<math>(0)-(2)</math>
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<math>0-2=\boxed{\textbf{(A)}\ {-}2}.</math>
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 +
~ghfhgvghj10
  
 
== See Also ==
 
== See Also ==

Revision as of 14:53, 18 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 $x\diamondsuit y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)?\] $\textbf{(A)}\ {-}2 \qquad\textbf{(B)}\ {-}1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 2$

Solution

We have \[(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)=(1\diamondsuit1)-(1\diamondsuit3)=0-2=\boxed{\textbf{(A)}\ {-}2}.\] ~MRENTHUSIASM

Solution 2 (Substiution)

$(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)$

$(|1-|2-3||)-(|1-2|-3|)$

$(|1-|-1||)-(|-1|-3|)$

$(|1-1|)-(|1-3|)$

$(|0|)-(|-2|)$

$(0)-(2)$

$0-2=\boxed{\textbf{(A)}\ {-}2}.$

~ghfhgvghj10

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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