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

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

Revision as of 20:37, 13 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} ~

2021 AMC 10B (Problems • Answer Key • Resources)
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