AMC 10 Step by Step
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2021 AMC 10A Problem 1

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2021 AMC 10A Problem 1AMC 10 Step by Step
Name:                             Date: September 13, 2026
Exponents, Logarithms & Radicals · difficulty 1 of 5 · about 1 min · position P1-10

What is the value of (22−2)−(32−3)+(42−4)?(2^2-2)-(3^2-3)+(4^2-4)?

  1. A)

    11

  2. B)

    22

  3. C)

    55

  4. D)

    88

  5. E)

    1212

Answer

D

Key insight

Evaluate each parenthesis separately: 2, 6, 12, then combine with the given signs to get 2 - 6 + 12 = 8.

Solution

Work inside each pair of parentheses first:
22−2=4−2=2,32−3=9−3=6,42−4=16−4=12.2^2 - 2 = 4 - 2 = 2,\qquad 3^2 - 3 = 9 - 3 = 6,\qquad 4^2 - 4 = 16 - 4 = 12.
Now combine them with the signs given in the expression:
2−6+12=8.2 - 6 + 12 = 8.

The answer is (D) 8\boxed{\textbf{(D)}\ 8}.

Why this works

Parentheses are evaluated before the operations between them, so a messy-looking expression becomes three small numbers. Notice each term is n2−n=n(n−1)n^2 - n = n(n-1), the product of consecutive integers (1⋅21\cdot 2, 2⋅32 \cdot 3, 3⋅43 \cdot 4), a pattern that shows up often on the AMC.

The trap

Applying the leading minus sign to only part of the second parenthesis, or subtracting the third term as well.

Common mistakes

  • Applying the leading minus sign to only part of the second parenthesis, or subtracting the third term as well.
  • Computing 22−2=22^2 - 2 = 2 correctly but then rushing and writing 42−4=84^2 - 4 = 8, which gives 44 instead of 88.

Techniques

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