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

(Instead of rounding pi to the nearest tenth or hundredth, we can also round it to the nearest whole.)
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==Video Solution by TheBeautyofMath==
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https://youtu.be/gLahuINjRzU
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~IceMatrix
  
 
{{AMC10 box|year=2021|ab=B|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2021|ab=B|before=First Problem|num-a=2}}

Revision as of 05:28, 19 February 2021

Problem

How many integer values of $x$ satisfy $|x|<3\pi$?

$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$

Solution 1

Since $3\pi$ is about $9.42$, we multiply 9 by 2 for the numbers from $1$ to $9$ and the numbers from $-1$ to $-9$ and add 1 to account for the zero to get $\boxed{\textbf{(D)}\ ~19}$~smarty101 and edited by Tony_Li2007

Solution 2

$3\pi \approx 9.4.$ There are two cases here.

When $x>0, |x|>0,$ and $x = |x|.$ So then $x<9.4$

When $x<0, |x|>0,$ and $x = -|x|.$ So then $-x<9.4$. Dividing by $-1$ and flipping the sign, we get $x>-9.4.$

From case 1 and 2, we know that $-9.4 < x < 9.4$. Since $x$ is an integer, we must have $x$ between $-9$ and $9$. There are a total of \[9-(-9) + 1 = \boxed{\textbf{(D)}\ ~19} \text{ integers}.\]

-PureSwag

Solution 3

$|x|<3\pi$ $\iff$ $-3\pi<x<3\pi$. Since $\pi$ is approximately $3.14$, $3\pi$ is approximately $9.42$. We are trying to solve for $-9.42<x<9.42$, where $x\in\mathbb{Z}$. Hence, $-9.42<x<9.42$ $\implies$ $-9\leq x\leq9$, for $x\in\mathbb{Z}$. The number of integer values of $x$ is $9-(-9)+1=19$. Therefore, the answer is $\boxed{\textbf{(D)}19}$.

~ {TSun} ~

Solution 4

Looking at the problem, we see that instead of directly saying $x$, we see that it is $|x|.$ That means all the possible values of $x$ in this case are positive and negative. Rounding $\pi$ to $3$ we get $3(3)=9.$ There are $9$ positive solutions and $9$ negative solutions. $9+9=18.$ But what about zero? Even though zero is neither negative nor positive, but we still need to add it into the solution. Hence, the answer is $9+9+1=18+1=\boxed{\textbf{(D)}19}$.

~DuoDuoling0

Video Solution 1

https://youtu.be/Hv9bQF5x1yQ

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/gLahuINjRzU

~IceMatrix

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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All AMC 10 Problems and Solutions
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