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

Revision as of 00:26, 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.

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

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

$\pi$ is approximately $\dfrac{22}{7}$ . So $7\pi = 7 \cdot \dfrac{22}{7} = 22$ (Keep on going)

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2024 AMC 10B Problems/Problem 3|2024 AMC 10B #3]] and [[2024 AMC 12B Problems/Problem 3|2024 AMC 12B #3]]}}
+==Problem==
+For how many integer values of <math>x</math> is <math>|2x| \leq 7 \pi</math>
+<math>\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 19 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 21</math>
+==Solution 1==
+<math>\pi</math> is approximately <math>\dfrac{22}{7}</math>. So <math>7\pi = 7 \cdot \dfrac{22}{7} = 22</math>
+(Keep on going)
+==See also==
+{{AMC10 box|year=2024|ab=B|num-b=2|num-a=4}}
+{{AMC12 box|year=2024|ab=B|num-b=2|num-a=4}}
+{{MAA Notice}}