Difference between revisions of "2021 AMC 12B Problems/Problem 1"

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{{duplicate|[[2021 AMC 10B Problems#Problem 1|2021 AMC 10B #1]] and [[2021 AMC 12B Problems#Problem 1|2021 AMC 12B #1]]}}
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==Problem==
 
==Problem==
  
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==Solution 1==
 
==Solution 1==
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
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==Solution 2==
 
==Solution 2==
 
<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>.
 
<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>
 
<br><br>
 
~ {TSun} ~
 
~ {TSun} ~
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==Solution 3==
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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 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>
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-PureSwag
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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
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==Video Solution by savannahsolver==
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https://youtu.be/Hv9bQF5x1yQ
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~savannahsolver
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==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==
 
https://youtube.com/watch?v=qpvS2PVkI8A
 
https://youtube.com/watch?v=qpvS2PVkI8A
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==Video Solution by TheBeautyofMath==
 
==Video Solution by TheBeautyofMath==
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https://youtu.be/gLahuINjRzU
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https://youtu.be/EMzdnr1nZcE
 
https://youtu.be/EMzdnr1nZcE
  
 
~IceMatrix
 
~IceMatrix
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==Video Solution by Interstigation==
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https://youtu.be/DvpN56Ob6Zw
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-Interstigation
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==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|before=First Problem|num-a=2}}
 
{{AMC12 box|year=2021|ab=B|before=First Problem|num-a=2}}
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{{AMC10 box|year=2021|ab=B|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 07:57, 2 March 2021

The following problem is from both the 2021 AMC 10B #1 and 2021 AMC 12B #1, so both problems redirect to this page.

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

$|x|<3\pi$ $\iff$ $-3\pi<x<3\pi$. 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 3

$3\pi \approx 9.4.$ 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 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 by savannahsolver

https://youtu.be/Hv9bQF5x1yQ

~savannahsolver

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=qpvS2PVkI8A

Video Solution by OmegaLearn (Basic Computation)

https://youtu.be/_C4ceJn6Iaw

Video Solution by Hawk Math

https://www.youtube.com/watch?v=VzwxbsuSQ80

Video Solution by TheBeautyofMath

https://youtu.be/gLahuINjRzU

https://youtu.be/EMzdnr1nZcE

~IceMatrix

Video Solution by Interstigation

https://youtu.be/DvpN56Ob6Zw

-Interstigation

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2021 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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