2020 AMC 10B Problem 7

1. An even multiple of $3$ is a number divisible by both $2$ and $3$, that is, a multiple of $6$.
2. Suppose a perfect square $m \times m$ is a multiple of $6$. Then $m$ itself must be even: if $m$ were odd, then $m \times m$ would be odd. In the same way $m$ must be a multiple of $3$, because $3$ is prime, so it can only divide $m \times m$ if it divides $m$. So $m$ is a multiple of $6$.
3. Now list the squares of multiples of $6$ and stop when they pass $2020$:

1. The first seven are below $2020$. The eighth, $2304$, is too big, and the squares only get bigger after that.

So there are $7$ such squares.

The answer is $\boxed{\textbf{(A)}\ 7}$.

If a prime divides a perfect square, it divides the number being squared. That turns "squares that are multiples of $6$" into "squares of multiples of $6$", which are easy to list. You can also shortcut the list: the squares are $36 \times 1, 36 \times 4, 36 \times 9, \ldots$, and $36 \times 49 = 1764 < 2020 < 2304 = 36 \times 64$.

An even multiple of $3$ is a multiple of $6$. If a perfect square $m^2$ is divisible by $2$ and by $3$, then $m$ itself is divisible by $2$ and $3$, so $m = 6k$ and $m^2 = 36k^2$.

We need $36k^2 < 2020$, i.e. $k^2 < \tfrac{2020}{36} \approx 56.1$. Since $7^2 = 49 < 56.1 < 64 = 8^2$, the valid values are $k = 1, 2, \dots, 7$.

Check the boundary: $36 \cdot 49 = 1764 < 2020$ and $36 \cdot 64 = 2304 > 2020$.

The answer is $\boxed{\textbf{(A)}\ 7}$.