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

(There is another solution for this. Instead of rounding pi to the nearest tenth or hundredth, we can round it to the nearest whole number.)
(Instead of rounding pi to the nearest tenth or hundredth, we can also round it to the nearest whole.)
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==Solution 4==
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Looking at the problem, we see that instead of directly saying <math>x</math>, we see that it is <math>|x|.</math> That means all the possible values of <math>x</math> in this case are positive and negative. Rounding <math>\pi</math> to <math>3</math> we get <math>3(3)=9.</math> There are <math>9</math> positive solutions and <math>9</math> negative solutions. <math>9+9=18.</math> 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 <math>9+9+1=18+1=\boxed{\textbf{(D)}19}</math>.
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~DuoDuoling0
  
 
==Video Solution 1==
 
==Video Solution 1==
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~savannahsolver
 
~savannahsolver
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==Solution 4==
 
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Looking at the problem, we see that instead of directly saying <math>x</math>, we see that it is <math>|x|.</math> That means all the possible values of <math>x</math> in this case are positive and negative. Rounding <math>\pi</math> to <math>3</math> we get <math>3(3)=9.</math> There are <math>9</math> positive solutions and <math>9</math> negative solutions. <math>9+9=18.</math> 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 <math>9+9+1=18+1=\boxed{\textbf{(D)}19}</math>.
 
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~DuoDuoling0
 
  
 
{{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 20:36, 15 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

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