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

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==Solution==
 
==Solution==
 
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
 
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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==Solution 2==
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<math>3\pi \approx 9.4.</math> There are two cases here.
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When <math>x>0, |x|>0,</math> and <math>x = |x|.</math> So then <math>x<9.4</math>
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When <math>x<0, |x|>0,</math> and <math>x = -|x|.</math> So then <math>-x<9.4</math>. Dividing by <math>-1</math> and flipping the sign, we get <math>x>-9.4.</math>
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From case 1 and 2, we need <math>-9.4 < x < 9.4</math>. Since <math>x</math> is an integer, we must have <math>x</math> between <math>-9</math> and <math>9</math>. There are a total of <cmath>9-(-9) + 1 = \boxed{\textbf{(D)}\ ~19} \text{ integers}.</cmath>
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-PureSwag

Revision as of 18:04, 11 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

Since $3\pi$ is about $9.42$, we multiply 9 by 2 and add 1 to get $\boxed{\textbf{(D)}\ ~19}$~smarty101

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 need $-9.4 < x < 9.4$. 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

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