# 2020 AMC 10B Problems/Problem 7

## 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 $\boxed{\textbf{(A) }7}$ valid answers. ~airbus-a321, November 2023

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~savannahsolver

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~ pi_is_3.14

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