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

Revision as of 20:42, 15 February 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

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

-happykeeper

Solution 2

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

Solution 3

You can just directly calculate them. Simplifying $(2^2-2)-(3^2-3)+(4^2-4),$ we get $(4-2)-(9-3)+(16-4),$ which simplifies to $2-6+12=8,$ so the answer is $\boxed{\textbf{(D)}8}.$

~DuoDuoling0

Video Solution #0(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 #1(Quick Computation)

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

Video Solution (Arithmetic Computation)

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

~ pi_is_3.14

Video Solution (Simple)

https://youtu.be/cAMg15KhK6E

Education, the Study of everything

Video Solution 5

https://youtu.be/4dkzuRHJieQ

~savannahsolver

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.

@@ Line 17: / Line 17: @@
 \end{align*}</cmath>
 ~MRENTHUSIASM
+==Solution 3==
+You can just directly calculate them. Simplifying <cmath>(2^2-2)-(3^2-3)+(4^2-4),</cmath> we get <cmath>(4-2)-(9-3)+(16-4),</cmath> which simplifies to <cmath>2-6+12=8,</cmath> so the answer is <math>\boxed{\textbf{(D)}8}.</math>
+~DuoDuoling0
 ==Video Solution #0(Very Very Quick Computation) ==

Page

Toolbox

Search