AMC 10 Step by Step

2022 AMC 10B Problem 1

P1-10~1 minPrint

Define x⋄yx\diamond y to be ∣x−y∣|x-y| for all real numbers xx and y.y. What is the value of (1⋄(2⋄3))−((1⋄2)⋄3)?(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?

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Key insight

The brackets decide the order. Work each half from the inside out, and do not expect the two groupings to give the same result.

Solution

The symbol ⋄\diamond just means "take the distance between the two numbers": x⋄y=∣x−y∣x \diamond y = |x - y|. The brackets tell you which pair to work out first, so do the innermost bracket first in each half.

  1. Left half, 1⋄(2⋄3)1 \diamond (2 \diamond 3). Inside first: 2⋄3=∣2−3∣=12 \diamond 3 = |2 - 3| = 1. Then 1⋄1=∣1−1∣=01 \diamond 1 = |1 - 1| = 0.
  2. Right half, (1⋄2)⋄3(1 \diamond 2) \diamond 3. Inside first: 1⋄2=∣1−2∣=11 \diamond 2 = |1 - 2| = 1. Then 1⋄3=∣1−3∣=21 \diamond 3 = |1 - 3| = 2.
  3. Subtract. 0−2=−20 - 2 = -2.

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

Why this works

With ordinary addition, (a+b)+c(a + b) + c and a+(b+c)a + (b + c) are always equal, so it is tempting to think brackets never matter. For a made-up operation like "distance between", they can matter a lot, and this problem is built to show that. When a question invents a new operation, follow the definition step by step and do not borrow rules from ordinary arithmetic.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Work from the inside out.

First half: 2⋄3=∣2−3∣=12\diamond 3 = |2-3| = 1, then 1⋄1=∣1−1∣=01\diamond 1 = |1-1| = 0.

Second half: 1⋄2=∣1−2∣=11\diamond 2 = |1-2| = 1, then 1⋄3=∣1−3∣=21\diamond 3 = |1-3| = 2.

The difference is 0−2=−20 - 2 = -2.

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

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.