2021 AMC 10A Problem 1

P1-10~1 minPrint

Exponents, Logarithms & Radicals Set up the equation/formula and compute; no special trick needed

What is the value of $(2^2-2)-(3^2-3)+(4^2-4)?$

Show solution without answering

Key insight

Work out each group in parentheses first. They give 2, 6 and 12, and then 2 - 6 + 12 = 8.

Solution

Parentheses always come first, so work out each group on its own and only then combine them.

1. First group: $2^2 - 2 = 4 - 2 = 2$ .
2. Second group: $3^2 - 3 = 9 - 3 = 6$ .
3. Third group: $4^2 - 4 = 16 - 4 = 12$ .
4. Now put the three results back with the signs that sit between the groups. The second group is subtracted and the third is added:
$2 - 6 + 12 = 8.$

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

Why this works

A long expression becomes easy once you break it into small pieces and handle one piece at a time. The minus sign in front of the second group belongs to that group only; the third group has its own plus sign.

Notice also that each group has the form $n^2 - n = n \times (n - 1)$ : the results $2 = 2 \times 1$ , $6 = 3 \times 2$ and $12 = 4 \times 3$ are products of two neighbouring whole numbers. That pattern comes up often on the AMC.

Concise solution

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

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

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

Review in the AoPS books

Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.4)
Introduction to Algebra, Chapter 19: Exponents and Logarithms (§19.4)