2021 AMC 12B Problems/Problem 1

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

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Video Solution by savannahsolver
8 Video Solution by Punxsutawney Phil
9 Video Solution by OmegaLearn (Basic Computation)
10 Video Solution by Hawk Math
11 Video Solution by TheBeautyofMath
12 Video Solution by Interstigation
13 See Also

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 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

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

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

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