2020 AMC 10B Problems/Problem 7

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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 1

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

Solution 2

A even multiple square of $3$ can be represented by $3^2 \cdot 2^2 \cdot x^2$, where $3^2$ is the multiple or $3$ and $2^2$ makes it even. Simplifying we have $36^2 \cdot x^2$. We can divide $2020$ by $36$ (floor) and get $56$ see the result. We can then see that there are $7$ different values for $x$. It can't be larger or else $x^2 > 56$. And thus $\boxed{\textbf{(A) }7}$

~ Wiselion

Solution 3

It can be seen that the problem is just asking for squares that are multiples of six. Thus, all squares of multiples of six can be listed out: $6^2$, $12^2$, $18^2$, $24^2$, $30^2$, $36^2$, and $42^2$. $48^2$=$2196 > 2020$. There are seven valid answers. ~airbus-a321, November 2023

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

https://youtu.be/OHR_6U686Qg


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

~savannahsolver

Video Solution by OmegaLearn

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
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All AMC 10 Problems and Solutions

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