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 Video Solution (Just 1 min!)
14 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 $3\pi\approx9.42$ , 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

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

Page

Toolbox

Search