Difference between revisions of "2024 AMC 10B Problems/Problem 3"
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Use the fact that <math>\pi \approx \dfrac{22}{7}</math>. Simplifying this gives <math>|2x| \leq 22</math>, which leads to <math>|x| \leq 11</math>. Now, all we have to do is count the number of possibilities for <math>x</math>, which is just <math>11 + 11 - 1 = \boxed{21}</math>. | Use the fact that <math>\pi \approx \dfrac{22}{7}</math>. Simplifying this gives <math>|2x| \leq 22</math>, which leads to <math>|x| \leq 11</math>. Now, all we have to do is count the number of possibilities for <math>x</math>, which is just <math>11 + 11 - 1 = \boxed{21}</math>. |
Revision as of 10:50, 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.
Contents
[hide]Problem
For how many integer values of is
Solution 1
is slightly less than . So The inequality expands to be . We find that can take the integer values between and inclusive. There are such values.
Note that if you did not know whether was greater than or less than , then you might perform casework. In the case that , the valid solutions are between and inclusive: solutions. Since, is not an answer choice, we can be confident that , and that 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 , you can use the extremely sophisticated method of just dividing via long division. Once you get to you realise that you don't need to divide further since when rounded to 4 decimal places.Therefore, you do not include and 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 . Simplifying this gives , which leads to . Now, all we have to do is count the number of possibilities for , which is just .
-jb2015007
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/DIl3rLQQkQQ?feature=shared
~ Pi Academy
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.