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

Revision as of 21:59, 5 June 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 $\begin{align*} (2^2-2)-(3^2-3)+(4^2-4) &= 2(2-1)-3(3-1)+4(4-1) \\ &= 2(1)-3(2)+4(3) \\ &= 2-6+12 \\ &= \boxed{\textbf{(D) } 8}. \end{align*}$ ~MRENTHUSIASM

Video Solution 1

https://youtu.be/slVBYmcDMOI

~TheLearningRoyal

2021 AMC 10A (Problems • Answer Key • Resources)
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 19: / Line 19: @@
 ~MRENTHUSIASM
-==Video Solution 8 (Problems 1-3)==
+==Video Solution 1==
-https://youtu.be/CupJpUzKPB0
+https://youtu.be/slVBYmcDMOI
-~MathWithPi
+~TheLearningRoyal
 == See Also ==
 {{AMC10 box|year=2021|ab=A|before=First Problem|num-a=2}}
 {{MAA Notice}}

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