Difference between revisions of "2024 AMC 10B Problems/Problem 3"
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==Problem== | ==Problem== | ||
For how many integer values of <math>x</math> is <math>|2x| \leq 7 \pi</math> | 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> | ||
+ | |||
+ | |||
+ | [ONLY FOR CERTAIN CHINESE TESTPAPERS] | ||
+ | |||
+ | For how many integer values of <math>x</math> is <math>|2x| \leq 6 \pi</math> | ||
<math>\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 19 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 21</math> | <math>\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 19 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 21</math> | ||
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~numerophile | ~numerophile | ||
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+ | Test advice: If you are in the test and do not know if <math>\frac{22}{7}</math> is bigger or smaller than <math>\pi</math>, you can use the extremely sophisticated method of just dividing <math>\dfrac{22}{7}</math> via long division. Once you get to <math>3.142</math> you realise that you don't need to divide further since <math>\pi = 3.1416</math> when rounded to 4 decimal places.Therefore, you do not include <math>11</math> and <math>-11</math> and the answer is 21. | ||
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+ | ~Rosiefork (first time using Latex)(and a complete noob) | ||
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+ | ==Solution 2== | ||
+ | [THIS SOLUTION DOES NOT WORK, PLEASE REFER TO SOLUTION 1] | ||
+ | |||
+ | 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>. | ||
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+ | -jb2015007 | ||
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+ | ==Solution 3== | ||
+ | <math>7\pi</math> is incredibly close to <math>22</math>, but doesn't reach it. This can be both computed by using <math>\pi\approx3.142\implies7\cdot3.142=21.994<22</math> or assumed. Therefore, including both positive and negative values, the answer is <math>\{-10,-9,...,9,10\}\implies\boxed{\text{(E) }21}</math>. ~Tacos_are_yummy_1 | ||
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+ | ==Solution 4== | ||
+ | [ONLY FOR CERTAIN CHINESE TESTPAPERS] | ||
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+ | Use the fact that <math>\pi \approx 3.14</math>, and thus you can get <math>6\pi \approx 18.84</math>. We could easily see that the answer is <math>\{-9,-8,...,8,9\}\implies\boxed{\text{(C) }19}</math> | ||
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+ | ~RULE101 | ||
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+ | ==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 | ||
+ | |||
+ | == Video Solution by Daily Dose of Math == | ||
+ | |||
+ | https://youtu.be/fJ2WqG-pchY | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
==See also== | ==See also== |
Revision as of 12:03, 16 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
[ONLY FOR CERTAIN CHINESE TESTPAPERS]
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 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
Solution 3
is incredibly close to , but doesn't reach it. This can be both computed by using or assumed. Therefore, including both positive and negative values, the answer is . ~Tacos_are_yummy_1
Solution 4
[ONLY FOR CERTAIN CHINESE TESTPAPERS]
Use the fact that , and thus you can get . We could easily see that the answer is
~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
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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.