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Difference between revisions of "2024 AMC 10B Problems/Problem 3"

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{{duplicate|[[2024 AMC 10B Problems/Problem 3|2024 AMC 10B #3]] and [[2024 AMC 12B Problems/Problem 3|2024 AMC 12B #3]]}}
  
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==Problem==
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For how many integer values of <math>x</math> is <math>|2x| \leq 7 \pi</math>
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<math>\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 19 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 21</math>
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[ONLY FOR CERTAIN CHINESE TESTPAPERS]
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For how many integer values of <math>x</math> is <math>|2x| \leq 6 \pi</math>
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<math>\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 19 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 21</math>
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==Solution 1==
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<math>\pi = 3.14159\dots</math> is slightly less than <math>\dfrac{22}{7} = 3.\overline{142857}</math>. So <math>7\pi \approx 21.9</math>
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The inequality expands to be <math>-21.9 \le 2x \le 21.9</math>. We find that <math>x</math> can take the integer values between <math>-10</math> and <math>10</math> inclusive. There are <math>\boxed{\text{E. }21}</math> such values.
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Note that if you did not know whether <math>\pi</math> was greater than or less than <math>\dfrac{22}{7}</math>, then you might perform casework. In the case that <math>\pi > \dfrac{22}{7}</math>, the valid solutions are between <math>-11</math> and <math>11</math> inclusive: <math>23</math> solutions. Since, <math>23</math> is not an answer choice, we can be confident that <math>\pi < \dfrac{22}{7}</math>, and that <math>\boxed{\text{E. } 21}</math> is the correct answer.
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~numerophile
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Test advice: If you are in the test and do not know if <math>\frac{22}{7}</math> is bigger or smaller than <math>\pi</math>, you can use the extremely sophisticated method of just dividing <math>\dfrac{22}{7}</math> via long division. Once you get to <math>3.142</math> you realise that you don't need to divide further since <math>\pi = 3.1416</math> when rounded to 4 decimal places.Therefore, you do not include <math>11</math> and <math>-11</math> and the answer is 21.
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~Rosiefork (first time using Latex)(and a complete noob)
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==Solution 3==
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<math>7\pi</math> is incredibly close to <math>22</math>, but doesn't reach it. This can be both computed by using <math>\pi\approx3.142\implies7\cdot3.142=21.994<22</math> or assumed. Therefore, including both positive and negative values, the answer is <math>\{-10,-9,...,9,10\}\implies\boxed{\text{(E) }21}</math>. ~Tacos_are_yummy_1
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==Solution 4==
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[ONLY FOR CERTAIN CHINESE TESTPAPERS]
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Use the fact that <math>\pi \approx 3.14</math>, and thus you can get <math>6\pi \approx 18.84</math>. We could easily see that the answer is <math>\{-9,-8,...,8,9\}\implies\boxed{\text{(C) }19}</math>
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~RULE101
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==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
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https://youtu.be/DIl3rLQQkQQ?feature=shared
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~ Pi Academy
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=24EZaeAThuE
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== Video Solution by Daily Dose of Math ==
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https://youtu.be/fJ2WqG-pchY
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~Thesmartgreekmathdude
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==See also==
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{{AMC10 box|year=2024|ab=B|num-b=2|num-a=4}}
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{{AMC12 box|year=2024|ab=B|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 12:14, 28 November 2024

The following problem is from both the 2024 AMC 10B #3 and 2024 AMC 12B #3, so both problems redirect to this page.

Problem

For how many integer values of $x$ is $|2x| \leq 7 \pi$

$\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 19 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 21$


[ONLY FOR CERTAIN CHINESE TESTPAPERS]

For how many integer values of $x$ is $|2x| \leq 6 \pi$

$\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 19 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 21$

Solution 1

$\pi = 3.14159\dots$ is slightly less than $\dfrac{22}{7} = 3.\overline{142857}$. So $7\pi \approx 21.9$ The inequality expands to be $-21.9 \le 2x \le 21.9$. We find that $x$ can take the integer values between $-10$ and $10$ inclusive. There are $\boxed{\text{E. }21}$ such values.

Note that if you did not know whether $\pi$ was greater than or less than $\dfrac{22}{7}$, then you might perform casework. In the case that $\pi > \dfrac{22}{7}$, the valid solutions are between $-11$ and $11$ inclusive: $23$ solutions. Since, $23$ is not an answer choice, we can be confident that $\pi < \dfrac{22}{7}$, and that $\boxed{\text{E. } 21}$ is the correct answer.

~numerophile

Test advice: If you are in the test and do not know if $\frac{22}{7}$ is bigger or smaller than $\pi$, you can use the extremely sophisticated method of just dividing $\dfrac{22}{7}$ via long division. Once you get to $3.142$ you realise that you don't need to divide further since $\pi = 3.1416$ when rounded to 4 decimal places.Therefore, you do not include $11$ and $-11$ and the answer is 21.

~Rosiefork (first time using Latex)(and a complete noob)

Solution 3

$7\pi$ is incredibly close to $22$, but doesn't reach it. This can be both computed by using $\pi\approx3.142\implies7\cdot3.142=21.994<22$ or assumed. Therefore, including both positive and negative values, the answer is $\{-10,-9,...,9,10\}\implies\boxed{\text{(E) }21}$. ~Tacos_are_yummy_1

Solution 4

[ONLY FOR CERTAIN CHINESE TESTPAPERS]

Use the fact that $\pi \approx 3.14$, and thus you can get $6\pi \approx 18.84$. We could easily see that the answer is $\{-9,-8,...,8,9\}\implies\boxed{\text{(C) }19}$

~RULE101

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/DIl3rLQQkQQ?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

Video Solution by Daily Dose of Math

https://youtu.be/fJ2WqG-pchY

~Thesmartgreekmathdude

See also

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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
2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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

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

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