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 $|2x| \leq 7 \pi$

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


[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 2

[THIS SOLUTION DOES NOT WORK, PLEASE REFER TO 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

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

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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