Difference between revisions of "2021 AMC 10B Problems/Problem 1"
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Revision as of 06:35, 2 March 2021
Contents
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} ~
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
Video Solution 1
~savannahsolver
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by Interstigation
-Interstigation
See Also
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 |
2021 AMC 12B (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.