2021 AMC 10B Problems/Problem 1
Revision as of 19:02, 15 February 2021 by Savannahsolver (talk | contribs)
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} ~
Video Solution 1
~savannahsolver
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |