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

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&=\boxed{\textbf{(A)}\ {-}2}.
 
&=\boxed{\textbf{(A)}\ {-}2}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
~MRENTHUSIASM & ghfhgvghj10
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~MRENTHUSIASM
  
 
== See Also ==
 
== See Also ==

Revision as of 21:42, 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 \begin{align*} (1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3) &= |1-|2-3|| - ||1-2|-3| \\ &= |1-1| - |1-3| \\ &= 0-2 \\ &=\boxed{\textbf{(A)}\ {-}2}. \end{align*} ~MRENTHUSIASM

See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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