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

Revision as of 04:34, 2 March 2021

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 and add 1 to get $\boxed{\textbf{(D)}\ ~19}$ ~smarty101

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} ~

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/EMzdnr1nZcE

~IceMatrix

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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 ==See Also==
 {{AMC12 box|year=2021|ab=B|before=First Problem|num-a=2}}
+{{AMC10 box|year=2021|ab=B|before=First Problem|num-a=2}}
 {{MAA Notice}}

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