2022 AMC 10B Problems/Problem 1: Difference between revisions

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

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, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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

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