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 | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/OHR_6U686Qg | ||
| + | |||
| + | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2020|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2020|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:14, 7 February 2020
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
~IceMatrix
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. 
