Difference between revisions of "2020 AMC 10B Problems/Problem 7"

(Video Solution)
(Video Solution)
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~savannahsolver
 
~savannahsolver
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== Video Solution ==
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https://youtu.be/ZhAZ1oPe5Ds?t=2241
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~ pi_is_3.14
  
 
==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 21:20, 17 January 2021

Problem

How many positive even multiples of $3$ less than $2020$ are perfect squares?

$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\  9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$

Solution

Any even multiple of $3$ is a multiple of $6$, so we need to find multiples of $6$ that are perfect squares and less than $2020$. Any solution that we want will be in the form $(6n)^2$, where $n$ is a positive integer. The smallest possible value is at $n=1$, and the largest is at $n=7$ (where the expression equals $1764$). Therefore, there are a total of $\boxed{\textbf{(A)}\ 7}$ possible numbers.-PCChess

Video Solution

Check It Out! Short & Straight-Forward Solution: Education, The Study of Everything

https://www.youtube.com/watch?v=igjvQv-TCGE


https://youtu.be/OHR_6U686Qg

~IceMatrix

https://youtu.be/5cDMRWNrH-U

~savannahsolver

Video Solution

https://youtu.be/ZhAZ1oPe5Ds?t=2241

~ pi_is_3.14

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
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

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