Difference between revisions of "2021 AMC 10B Problems/Problem 1"
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When <math>x<0, |x|>0,</math> and <math>x = -|x|.</math> So then <math>-x<9.4</math>. Dividing by <math>-1</math> and flipping the sign, we get <math>x>-9.4.</math> | When <math>x<0, |x|>0,</math> and <math>x = -|x|.</math> So then <math>-x<9.4</math>. Dividing by <math>-1</math> and flipping the sign, we get <math>x>-9.4.</math> | ||
− | From case 1 and 2, we | + | From case 1 and 2, we know that <math>-9.4 < x < 9.4</math>. Since <math>x</math> is an integer, we must have <math>x</math> between <math>-9</math> and <math>9</math>. There are a total of <cmath>9-(-9) + 1 = \boxed{\textbf{(D)}\ ~19} \text{ integers}.</cmath> |
-PureSwag | -PureSwag |
Revision as of 22:18, 12 February 2021
Contents
[hide]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
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) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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All AMC 10 Problems and Solutions |