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

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==Solution 1==
 
==Solution 1==
 
<math>\pi = 3.14159\cdots</math> is slightly less than <math>\dfrac{22}{7} = \3.\overline{142857}</math>. So <math>7\pi \approx 21.9</math>
 
<math>\pi = 3.14159\cdots</math> is slightly less than <math>\dfrac{22}{7} = \3.\overline{142857}</math>. So <math>7\pi \approx 21.9</math>
−
The inequality expands to be <math>-21.9 <= 2x <= 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{E. 21}</math> such values.
+
The inequality expands to be <math>-21.9 <= 2x <= 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.
  
−
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{E. 21}</math> is the correct answer.
+
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.
 +
 
 +
~numerophile
  
 
==See also==
 
==See also==

Revision as of 03:12, 14 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 $|2x| \leq 7 \pi$

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

Solution 1

$\pi = 3.14159\cdots$ is slightly less than $\dfrac{22}{7} = \3.\overline{142857}$ (Error compiling LaTeX. Unknown error_msg). So $7\pi \approx 21.9$ The inequality expands to be $-21.9 <= 2x <= 21.9$. 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

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

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