2020 AMC 10B Problem 7

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 $\boxed{\textbf{(A)}\ 7}$.

Divisibility conditions on a perfect square transfer to its root: if $p \mid m^2$ for a prime $p$, then $p \mid m$, so $p^2 \mid m^2$. This turns "squares that are multiples of $6$" into "squares of multiples of $6$," which are trivial to count by bounding.

• Counting multiples of 6 that are less than 2020 rather than squares, or forgetting that a square divisible by 6 needs 36.
• Counting squares that are multiples of $3$ only ($9k^2 < 2020$ gives $k \le 14$) and forgetting the evenness condition, or including $k = 8$ because $36 \cdot 64$ "looks" close.