Will you find all the perfect squares?

We can see that $24=2^3×3$ , so the first perfect square that is a multiple of $24$ would be $2^4×3^2=12^2=144$ .

Next would be $2^6×3^2=24^2=576$ and then $2^4×3^4=36^2=1296$ and so on.

We can see that $n^2$ would be a multiple of $24$ only when $n$ is a multiple of $12$ .

So we just have to find the multiples of $12$ less than $\sqrt{10^6}=10^3$ .

$\begin{aligned} 996 & = 12+(n-1)12 \\ 996 & = 12 n \\ \Rightarrow n & = \frac{996}{12}=\boxed{83}\end{aligned}$

Since $24=2^33$ the perfect square multiple of 24 must be of the form $2^43^2k^2 = 144k^2$ , where $k = 1,2,3,...$ . The largest $k$ , which is also the number of such multiples less than $10^6$ is given by:

$\begin{aligned} 144k^2 & < 10^6 \\ k & < \sqrt{\frac{10^6}{144}} = \frac{1000}{12} \\ \Rightarrow k & = \left \lfloor \frac{1000}{12} \right \rfloor = \left \lfloor 83.333 \right \rfloor = \boxed{83} \end{aligned}$

Since $24=2^{3} \times 3$ , then the least perfect square greater than 24 that is a multiple of 24 is $2^{4} \times 3^{2}=144$ . Thus, any perfect square that is divisible by 24 must be of the form $x^{2} \times 12^{2}$ . So we have, $12^{2}, 2^{2} \times 12^{2},...,n^{2} \times 12 ^{2}$ . The largest number n such that $n^{2}\times 12^{2} < 10^{6}$ is when $n=83$ . Hence, there are $\boxed{83}$ such numbers.

This is from AIME 2007 contest.