Difference between revisions of "2021 AMC 12B Problems/Problem 1"
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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]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
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==Solution 1== | ==Solution 1== | ||
− | Since <math>3\pi</math> is about <math>9.42</math>, we multiply 9 by 2 and add 1 to get <math> \boxed{\textbf{(D)}\ ~19} </math>~smarty101 | + | Since <math>3\pi</math> is about <math>9.42</math>, we multiply 9 by 2 for the numbers from <math>1</math> to <math>9</math> and the numbers from <math>-1</math> to <math>-9</math> and add 1 to account for the zero to get <math>\boxed{\textbf{(D)}\ ~19}</math>~smarty101 and edited by Tony_Li2007 |
+ | |||
==Solution 2== | ==Solution 2== | ||
<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>. | <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>. | ||
<br><br> | <br><br> | ||
~ {TSun} ~ | ~ {TSun} ~ | ||
+ | |||
+ | ==Solution 3== | ||
+ | <math>3\pi \approx 9.4.</math> There are two cases here. | ||
+ | |||
+ | When <math>x>0, |x|>0,</math> and <math>x = |x|.</math> So then <math>x<9.4</math> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | -PureSwag | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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>. | ||
+ | |||
+ | ~DuoDuoling0 | ||
+ | |||
+ | ==Solution 5== | ||
+ | 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== | ||
+ | https://youtu.be/Hv9bQF5x1yQ | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://youtube.com/watch?v=qpvS2PVkI8A | https://youtube.com/watch?v=qpvS2PVkI8A | ||
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==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/gLahuINjRzU | ||
+ | |||
https://youtu.be/EMzdnr1nZcE | https://youtu.be/EMzdnr1nZcE | ||
~IceMatrix | ~IceMatrix | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/DvpN56Ob6Zw | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|before=First Problem|num-a=2}} | {{AMC12 box|year=2021|ab=B|before=First Problem|num-a=2}} | ||
{{AMC10 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}} | {{MAA Notice}} |
Latest revision as of 23:14, 20 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.
Contents
- 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 satisfy ?
Solution 1
Since is about , we multiply 9 by 2 for the numbers from to and the numbers from to and add 1 to account for the zero to get ~smarty101 and edited by Tony_Li2007
Solution 2
. Since is approximately , is approximately . We are trying to solve for , where . Hence, , for . The number of integer values of is . Therefore, the answer is .
~ {TSun} ~
Solution 3
There are two cases here.
When and So then
When and So then . Dividing by and flipping the sign, we get
From case 1 and 2, we know that . Since is an integer, we must have between and . There are a total of -PureSwag
Solution 4
Looking at the problem, we see that instead of directly saying , we see that it is That means all the possible values of in this case are positive and negative. Rounding to we get There are positive solutions and negative solutions. 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 .
~DuoDuoling0
Solution 5
There are an odd number of integer solutions to this inequality since if any non-zero integer satisfies this inequality, then so does and we must also account for 0, which gives us the desired. Then, the answer is either or and since the answer is at least yielding
Video Solution by savannahsolver
~savannahsolver
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=qpvS2PVkI8A
Video Solution by OmegaLearn (Basic Computation)
Video Solution by Hawk Math
https://www.youtube.com/watch?v=VzwxbsuSQ80
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by Interstigation
~Interstigation
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.