Difference between revisions of "2021 AMC 10B Problems/Problem 1"
Tony li2007 (talk | contribs) (→Solution 1) |
|||
Line 21: | Line 21: | ||
<br><br> | <br><br> | ||
~ {TSun} ~ | ~ {TSun} ~ | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/Hv9bQF5x1yQ | ||
+ | |||
+ | ~savannahsolver | ||
{{AMC10 box|year=2021|ab=B|before=First Problem|num-a=2}} | {{AMC10 box|year=2021|ab=B|before=First Problem|num-a=2}} |
Revision as of 19:02, 15 February 2021
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 |