2021 AMC 10B Problems/Problem 1
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 ![\[9-(-9) + 1 = \boxed{\textbf{(D)}\ ~19} \text{ integers}.\]](http://latex.artofproblemsolving.com/e/a/9/ea9efa4d90ca73485400407f03c2a32b31698103.png)
-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} ~
Video Solution 1
~savannahsolver
Solution 4
Looking at the problem, we see that instead of directly saying , we see that it is 
That means all the possible values of in this case are positive and negative. Rounding to 
we get 
There are positive solutions and negative solutions. 
But what about zero? Even though zero is neither negative nor positive, but we still need to add it into the solution. Hence, the answer is 
.
~DuoDuoling0
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