2023 AMC 10A Problem 3

1. First decide which squares are divisible by $5$. Try a few: $5^2 = 25$ and $10^2 = 100$ are, but $6^2 = 36$ and $7^2 = 49$ are not. Since $5$ is prime, $5$ can only divide $n \times n$ if it divides $n$ itself. So we want the squares of $5, 10, 15, 20, \ldots$
2. Next find where the squares stop being less than $2023$. We know $40^2 = 1600$ and $45^2 = 2025$.
3. $2025$ is just above $2023$, so $45$ is not allowed. The last one that works is $40$.
4. List them: $5, 10, 15, 20, 25, 30, 35, 40$. Their squares $25, 100, 225, 400, 625, 900, 1225, 1600$ are all below $2023$ and all divisible by $5$.
5. That is $8$ squares.

The answer is $\boxed{\textbf{(A)}\ 8}$.

A prime number divides a square only when it divides the number being squared, so the question about squares turns into a question about multiples of $5$. The only tricky part is the cutoff: $45^2 = 2025$ misses $2023$ by just $2$, and that single case decides between (A) and (B). Always check the last case by hand.

Because $5$ is prime, $5$ divides $n^2$ exactly when $5$ divides $n$. So we need to count positive multiples of $5$ whose squares are below $2023$.

Since $44^2 = 1936 < 2023$ and $45^2 = 2025 > 2023$, the roots allowed are $n = 1, 2, \ldots, 44$. The multiples of $5$ in this range are $5, 10, 15, \ldots, 40$, which is $8$ numbers.

Their squares are $25, 100, 225, \ldots, 1600$, all below $2023$ and all divisible by $5$.

The answer is $\boxed{\textbf{(A)}\ 8}$.