Difference between revisions of "2020 AMC 10B Problems/Problem 7"
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Any even multiple of <math>3</math> is a multiple of <math>6</math>, so we need to find multiples of <math>6</math> that are perfect squares and less than <math>2020</math>. Any solution that we want will be in the form <math>(6n)^2</math>, where <math>n</math> is a positive integer. The smallest possible value is at <math>n=1</math>, and the largest is at <math>n=7</math> (where the expression equals <math>1764</math>). Therefore, there are a total of <math>\boxed{\textbf{(A)}\ 7}</math> possible numbers.-PCChess | Any even multiple of <math>3</math> is a multiple of <math>6</math>, so we need to find multiples of <math>6</math> that are perfect squares and less than <math>2020</math>. Any solution that we want will be in the form <math>(6n)^2</math>, where <math>n</math> is a positive integer. The smallest possible value is at <math>n=1</math>, and the largest is at <math>n=7</math> (where the expression equals <math>1764</math>). Therefore, there are a total of <math>\boxed{\textbf{(A)}\ 7}</math> possible numbers.-PCChess | ||
| − | ==Video Solution== | + | ==Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)== |
| − | Check It Out! Short & Straight-Forward Solution | + | https://www.youtube.com/watch?v=igjvQv-TCGE |
| − | Education, The Study of Everything | + | |
| + | Check It Out! Short & Straight-Forward Solution | ||
| + | ~Education, The Study of Everything | ||
| − | |||
| + | ==Video Solutions== | ||
https://youtu.be/OHR_6U686Qg | https://youtu.be/OHR_6U686Qg | ||
| − | |||
https://youtu.be/5cDMRWNrH-U | https://youtu.be/5cDMRWNrH-U | ||
Revision as of 12:41, 6 June 2023
Contents
Problem
How many positive even multiples of 
less than 
are perfect squares?
Solution
Any even multiple of is a multiple of 
, so we need to find multiples of that are perfect squares and less than . Any solution that we want will be in the form 
, where 
is a positive integer. The smallest possible value is at 
, and the largest is at 
(where the expression equals 
). Therefore, there are a total of 
possible numbers.-PCChess
Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)
https://www.youtube.com/watch?v=igjvQv-TCGE
Check It Out! Short & Straight-Forward Solution ~Education, The Study of Everything
Video Solutions
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/ZhAZ1oPe5Ds?t=2241
~ pi_is_3.14
See Also
| 2020 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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. 
