Difference between revisions of "2021 AMC 10B Problems/Problem 1"
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==Solution 1== | ==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 for the numbers from <math>1</math> to <math>9</math> and the numbers from <math>-1</math> to <math>-9</math> and add 1 to account for the zero to get <math> \boxed{\textbf{(D)}\ ~19} </math>~smarty101 and edited by Tony_Li2007 |
==Solution 2== | ==Solution 2== |
Revision as of 20:37, 13 February 2021
Contents
[hide]Problem
How many integer values of satisfy ?
Solution 1
Since is about , we multiply 9 by 2 for the numbers from to and the numbers from to and add 1 to account for the zero to get ~smarty101 and edited by Tony_Li2007
Solution 2
There are two cases here.
When and So then
When and So then . Dividing by and flipping the sign, we get
From case 1 and 2, we know that . Since is an integer, we must have between and . There are a total of
-PureSwag
Solution 3
. 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} ~
2021 AMC 10B (Problems • Answer Key • Resources) | ||
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