

Line 1: 
Line 1: 
−  #redirect [[2021 AMC 12B Problems/Problem 1Solution]]  +  #REDIRECT [[2021 AMC 12B Problems/Problem 1Solution]] 
−   
−  ==Problem==
 
−  How many integer values of <math>x</math> satisfy <math>x<3\pi</math>?
 
−   
−  <math>\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20</math>
 
−   
−  ==Solution 1==
 
−  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==
 
−  <math>3\pi \approx 9.4.</math> There are two cases here.
 
−   
−  When <math>x>0, x>0,</math> and <math>x = x.</math> So then <math>x<9.4</math>
 
−   
−  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>
 
−   
−  From case 1 and 2, we know that <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>
 
−   
−  PureSwag
 
−  ==Solution 3==
 
−  <math>x<3\pi</math> <math>\iff</math> <math>3\pi<x<3\pi</math>. Since <math>\pi</math> is approximately <math>3.14</math>, <math>3\pi</math> is approximately <math>9.42</math>. We are trying to solve for <math>9.42<x<9.42</math>, where <math>x\in\mathbb{Z}</math>. Hence, <math>9.42<x<9.42</math> <math>\implies</math> <math>9\leq x\leq9</math>, for <math>x\in\mathbb{Z}</math>. The number of integer values of <math>x</math> is <math>9(9)+1=19</math>. Therefore, the answer is <math>\boxed{\textbf{(D)}19}</math>.
 
−  <br><br>
 
−  ~ {TSun} ~
 
−  ==Solution 4==
 
−  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>.
 
−   
−  ~DuoDuoling0
 
−   
−  ==Video Solution 1==
 
−  https://youtu.be/Hv9bQF5x1yQ
 
−   
−  ~savannahsolver
 
−   
−  ==Video Solution by TheBeautyofMath==
 
−  https://youtu.be/gLahuINjRzU
 
−   
−  ~IceMatrix
 
−   
−  ==Video Solution by Interstigation==
 
−  https://youtu.be/DvpN56Ob6Zw
 
−   
−  Interstigation
 
−   
−  ==See Also==
 
−  {{AMC10 boxyear=2021ab=Bbefore=First Problemnuma=2}}
 
−  {{AMC12 boxyear=2021ab=Bbefore=First Problemnuma=2}}
 
−  {{MAA Notice}}
 