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Difference between revisions of "2021 AMC 10A Problems/Problem 1"

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&=2^2+7-3=2^2+4 \\
 
&=2^2+7-3=2^2+4 \\
 
&=4\cdot 2 \\
 
&=4\cdot 2 \\
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&=\boxed{\textbf{(D) } 8}.</cmath>
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&=\boxed{\textbf{(D) } 8}.
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\end{align*}</cmath>
 
-[[User:PureSwag|PureSwag]]
 
-[[User:PureSwag|PureSwag]]
  

Revision as of 20:47, 8 August 2021

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*} \left(2^2-2\right)-\left(3^2-3\right)+\left(4^2-4\right) &= 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

Solution 3 (Overkill, Just for Fun)

We have \begin{align*} (2^2-2)-(3^2-3)+(4^2-4) &= 2^2+4^2-3^2-2+3-4 \\ &=2^2+(4-3)(4+3)-3 \\ &=2^2+7-3=2^2+4 \\ &=4\cdot 2 \\ &=\boxed{\textbf{(D) } 8}. \end{align*} -PureSwag

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

See Also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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

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