2024 AMC 10B Problems/Problem 3

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

$\pi = 3.14159\dots$ 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 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 1

Use the fact that $\pi \approx \dfrac{22}{7}$ . Simplifying this gives $|2x| \leq 22$ , which leads to $|x| \leq 11$ . Now, all we have to do is count the number of possibilities for $x$ , which is just $11 + 11 - 1 = \boxed{21}$ .

-jb2015007

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

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

~ Pi Academy

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.

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2024 AMC 10B Problems/Problem 3

Contents

Problem

Solution 1

Solution 1

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

See also