Numbers

The prime factorization of $24$ is $2^3\times 3$ . Thus, each square must have at least $3$ factors of $2$ and $1$ factor of $3$ and its square root must have $2$ factors of $2$ and $1$ factor of $3$ . This means that each square is in the form $(12c)^2$ , where $c$ is a positive integer less than $\sqrt{10^6}$ . There are $\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{83}$ solutions.

$24=2^2×6$

$\sqrt {10^6}=10^3=1000$

1000÷2=500

$\lfloor \frac{500}{6}\rfloor= \boxed {83}$

24=2^3 \times 3

For a perfect square to be a multiple of 24, it has to be a multiple of 2^4 \times 3^4=144.

10^6\144 \approx 6944.44.

Now we need to find the number of perfect squares below 6944.44.

\sqrt{6944.44} \approx 83.33

\lfloor 83.33 \rfloor = \boxed{83}