2021 AMC 10A Problems/Problem 1

Problem

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

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$

Solution 1

$(4-2)-(9-3)+(16-4)=2-6+12=8.$ This corresponds to $\boxed{\textbf{(D) } 8}$

-happykeeper

Solution 2

We have $\begin{align*} (2^2-2)-(3^2-3)+(4^2-4) &= 2(2-1)-3(3-1)+4(4-1) \\ &= 2(1)-3(2)+4(3) \\ &= 2-6+12 \\ &= \boxed{\textbf{(D) } 8}. \end{align*}$ ~MRENTHUSIASM

Video Solution 6

https://youtu.be/4dkzuRHJieQ

~savannahsolver

Video Solution 7

https://youtu.be/50CThrk3RcM

~IceMatrix

Video Solution 8 (Problems 1-3)

https://youtu.be/CupJpUzKPB0

~MathWithPi

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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