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

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The 2021 AMC 12B will be held on February 10th, 2021. The problems will not be made public until 24 hours after that.
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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==
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How many integer values of <math>x</math> satisfy <math>|x|<3\pi</math>?
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<math>\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20</math>
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==Solution 1==
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Since <math>3\pi\approx9.42</math>, we multiply <math>9</math> by <math>2</math> for the integers from <math>1</math> to <math>9</math> and the integers from <math>-1</math> to <math>-9</math> and add <math>1</math> to account for <math>0</math> to get <math>\boxed{\textbf{(D)} ~19}</math>.
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~smarty101 and edited by Tony_Li2007 and sl_hc
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==Solution 2==
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<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>.
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~ {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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~[[User:PureSwag|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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==Solution 5==
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There are an odd number of integer solutions <math>x</math> to this inequality since if any non-zero integer <math>x</math> satisfies this inequality, then so does <math>-x,</math> and we must also account for <math>0,</math> which gives us the desired. Then, the answer is either <math>\textbf{(A)}</math> or <math>\textbf{(D)},</math> and since <math>3 \pi > 3 \cdot 3 > 9,</math> the answer is at least <math>9 \cdot 2 + 1 = 19,</math> yielding <math>\boxed{\textbf{(D)} ~19}.</math>
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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==
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https://youtube.com/watch?v=qpvS2PVkI8A
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== Video Solution by OmegaLearn (Basic Computation) ==
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https://youtu.be/_C4ceJn6Iaw
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==Video Solution by Hawk Math==
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https://www.youtube.com/watch?v=VzwxbsuSQ80
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==Video Solution by TheBeautyofMath==
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https://youtu.be/gLahuINjRzU
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https://youtu.be/EMzdnr1nZcE
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~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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==Video Solution (Just 1 min!)==
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https://youtu.be/9hN8QVAR748
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~Education, the Study of Everything
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==See Also==
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{{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}}
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{{MAA Notice}}

Latest revision as of 22:48, 18 July 2023

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 $3\pi\approx9.42$, we multiply $9$ by $2$ for the integers from $1$ to $9$ and the integers from $-1$ to $-9$ and add $1$ to account for $0$ to get $\boxed{\textbf{(D)} ~19}$.

~smarty101 and edited by Tony_Li2007 and sl_hc

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

Solution 5

There are an odd number of integer solutions $x$ to this inequality since if any non-zero integer $x$ satisfies this inequality, then so does $-x,$ and we must also account for $0,$ which gives us the desired. Then, the answer is either $\textbf{(A)}$ or $\textbf{(D)},$ and since $3 \pi > 3 \cdot 3 > 9,$ the answer is at least $9 \cdot 2 + 1 = 19,$ yielding $\boxed{\textbf{(D)} ~19}.$

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

Video Solution (Just 1 min!)

https://youtu.be/9hN8QVAR748

~Education, the Study of Everything

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

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