2021 AMC 10B Problems/Problem 1
Contents
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
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 2021 AMC 10A |
Followed by 2022 AMC 10A | |
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