# 2011 AMC 10B Problems/Problem 15

## Problem

Let $@$ denote the "averaged with" operation: $a @ b = \frac{a+b}{2}$. Which of the following distributive laws hold for all numbers $x, y,$ and $z$? $$\text{I. x @ (y + z) = (x @ y) + (x @ z)}$$ $$\text{II. x + (y @ z) = (x + y) @ (x + z)}$$ $$\text{III. x @ (y @ z) = (x @ y) @ (x @ z)}$$

$\textbf{(A)}\ \text{I only} \qquad\textbf{(B)}\ \text{II only} \qquad\textbf{(C)}\ \text{III only} \qquad\textbf{(D)}\ \text{I and III only} \qquad\textbf{(E)}\ \text{II and III only}$

## Solution

Simplify each operation and see which ones hold true.

\begin{align*} \text{I.} \qquad x @ (y + z) &= (x @ y) + (x @ z)\\ \frac{x+y+z}{2} &= \frac{x+y}{2} + \frac{x+z}{2}\\ \frac{x+y+z}{2} &\not= \frac{2x+y+z}{2} \end{align*}

\begin{align*} \text{II.} \qquad x + (y @ z) &= (x + y) @ (x + z)\\ x+ \frac{y+z}{2} &= \frac{2x+y+z}{2}\\ \frac{2x+y+z}{2} &= \frac{2x+y+z}{2} \end{align*}

\begin{align*} \text{III.} \qquad x @ (y @ z) &= (x @ y) @ (x @ z)\\ x @ \frac{y+z}{2} &= \frac{x+y}{2} @ \frac{x+z}{2}\\ \frac{2x+y+z}{4} &= \frac{2x+y+z}{4} \end{align*}

$\boxed{\textbf{(E)} \text{II and III only}}$

## Solution 2

Alternatively, substitute arbitrary values for $x$, $y$, and $z$ where $x\neq y\neq z$. For example, $x=1$, $y=2$, and $z=3$ give that only II and III work. $\boxed{\textbf{(E)} \text{II and III only}}$