2020 AMC 10B Problems/Problem 7

Revision as of 16:21, 6 April 2021 by Bakedpotato66 (talk | contribs) (Video Solution)

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png