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 1

https://youtu.be/slVBYmcDMOI

~TheLearningRoyal

2021 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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2021 AMC 10A Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

Video Solution 1

See Also