2021 AMC 10A Problems/Problem 1: Difference between revisions

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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 (Lightening Fast)

https://youtu.be/cAMg15KhK6E

~ Education, the Study of everything

Video Solution 2

https://youtu.be/cAMg15KhK6E

Video Solution 3 (Very Very Quick Computation)

https://www.youtube.com/watch?v=m0_UMI2mnZs&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=2 ~North America Mathematic Contest Go Go Go

Video Solution 4 (Quick Computation)

https://youtu.be/C3n2hgBhyXc?t=37 ~ThePuzzlr

Video Solution 5 (Arithmetic Computation)

https://youtu.be/0VvM_f-IDvE

~ pi_is_3.14

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

Video Solution 9

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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