What is the maximum power of 24 that can divide 124!?

24 = $$\ 2^3\times3$$

We need to see the maximum power of both.

Maximum power of 2 in 124! = $$\left[\dfrac{124}{2}\right]+\left[\dfrac{124}{4}\right]+\left[\dfrac{124}{8}\right]+\left[\dfrac{124}{16}\right]+\left[\dfrac{124}{32}\right]+\left[\dfrac{124}{64}\right]$$ = 119

So, the maximum power of $$2^3$$ in 124! is $$\left[\dfrac{119}{3}\right]=39$$

Now, let's check for 3: $$\left[\dfrac{124}{3}\right]+\left[\dfrac{124}{9}\right]+\left[\dfrac{124}{27}\right]+\left[\dfrac{124}{81}\right]=41+13+4+1=59$$

Since the maximum power that can be common for both $$2^3$$ and 3 is 39, the maximum power for 24 will also be 39.