# 2021 AMC 10A Problems/Problem 1

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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*} \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

## Solution 4 (When you have too much time)

Using the difference of squares, we have $$(2-\sqrt{2})(2+\sqrt{2}) - (3-\sqrt{3})(3+\sqrt{3}) + (4-2)(4+2).$$ Knowing that $\sqrt{2} \approx 1.41$ and $\sqrt{3} \approx 1.73,$ we get $$(2-\sqrt{2})(2+\sqrt{2}) - (3-\sqrt{3})(3+\sqrt{3}) + (4-2)(4+2) \approx 0.59\cdot 3.41 -1.26\cdot 4.73 + 2 \cdot 6 =8.0521,$$ which is closest to $\boxed{\textbf{(D) } 8}.$

## Video Solution 1 (Lightning Fast)

~ Education, the Study of everything

## 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

~ pi_is_3.14

~savannahsolver

~IceMatrix

~MathWithPi

## Video Solution 9

~TheLearningRoyal