Difference between revisions of "2021 AMC 12B Problems/Problem 1"
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==See Also== | ==See Also== | ||
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+ | {{AMC10 box|year=2021|ab=B|before=First Problem|num-a=2}} | ||
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Revision as of 05:34, 2 March 2021
Contents
Problem
How many integer values of satisfy ?
Solution 1
Since is about , we multiply 9 by 2 and add 1 to get ~smarty101
Solution 2
. 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 by Punxsutawney Phil
https://youtube.com/watch?v=qpvS2PVkI8A
Video Solution by OmegaLearn (Basic Computation)
Video Solution by Hawk Math
https://www.youtube.com/watch?v=VzwxbsuSQ80
Video Solution by TheBeautyofMath
~IceMatrix
See Also
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.