# 2022 AMC 10B Problems/Problem 1

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⚡️)

~Education, the Study of Everything

~~Hayabusa1