Difference between revisions of "2021 AMC 10B Problems/Problem 1"
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==Solution== | ==Solution== | ||
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 | ||
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+ | ==Solution 2== | ||
+ | <math>3\pi \approx 9.4.</math> There are two cases here. | ||
+ | |||
+ | When <math>x>0, |x|>0,</math> and <math>x = |x|.</math> So then <math>x<9.4</math> | ||
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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> | ||
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+ | From case 1 and 2, we need <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> | ||
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+ | -PureSwag |
Revision as of 19:04, 11 February 2021
Problem
How many integer values of satisfy
?
Solution
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 need . Since
is an integer, we must have
between
and
. There are a total of
-PureSwag