Printable

2022 AMC 10B Problem 1

2022 AMC 10B Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

Functions · difficulty 1 of 5 · about 1 min · position P1-10

Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of $(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?$

A)
${-}2$
B)
${-}1$
C)
$0$
D)
$1$
E)
$2$

Answer

Key insight

Evaluate the innermost diamond first in each half: 1◇(2◇3) = 1◇1 = 0 while (1◇2)◇3 = 1◇3 = 2.

Solution

Work from the inside out.

First half: $2\diamond 3 = |2-3| = 1$ , then $1\diamond 1 = |1-1| = 0$ .

Second half: $1\diamond 2 = |1-2| = 1$ , then $1\diamond 3 = |1-3| = 2$ .

The difference is $0 - 2 = -2$ .

The answer is $\boxed{\textbf{(A)}\ -2}$ .

Why this works

A "defined operation" is just a two-input function; the parentheses tell you exactly which pair to feed in first. The whole point of the problem is that $|x-y|$ is not associative, so the two groupings give different values. Whenever a custom operator appears, evaluate mechanically and never assume familiar laws (associativity, distributivity) carry over.

The trap

Assuming the operation is associative and answering 0, or dropping the absolute value and getting a sign wrong.

Common mistakes

Assuming the operation is associative and answering 0, or dropping the absolute value and getting a sign wrong.
Subtracting in the wrong order at the end and choosing (E) $2$ .

Techniques

Set up the equation/formula and compute; no special trick needed