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

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~DuoDuoling0
 
~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 0, 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)}}.</math>
  
 
==Video Solution by savannahsolver==
 
==Video Solution by savannahsolver==

Revision as of 11:59, 19 April 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

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 $\boxed{\textbf{(D)}}.$

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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