Difference between revisions of "2021 AMC 12B Problems/Problem 1"
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<math>\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20</math> | <math>\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Since <math>3\pi</math> is about <math>9.42</math>, we multiply 9 by 2 and add 1 to get <math> \boxed{\textbf{(D)}\ ~19} </math>~smarty101 | Since <math>3\pi</math> is about <math>9.42</math>, we multiply 9 by 2 and add 1 to get <math> \boxed{\textbf{(D)}\ ~19} </math>~smarty101 | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 12:03, 12 February 2021
Contents
Problem
How many integer values of 
satisfy 
?
Solution 1
Since 
is about 
, we multiply 9 by 2 and add 1 to get 
~smarty101
Solution 2
. Since 
is approximately 
, is approximately . We are trying to solve for 
, where 
. Hence, 
, for . The number of integer values of is 
. Therefore, the answer is 
.
~ {TSun} ~
Video Solution by OmegaLearn (Basic Computation)
Video Solution by Hawk Math
https://www.youtube.com/watch?v=VzwxbsuSQ80
See Also
| 2021 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 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. 
