Using the properties of exponents that (ab)^x = a^xb^x and (a^x)^y = a^(xy), we have
24^k = (2^3*3^1)^k = (2^3)^k*(3^1)^k = 2^(3k)*3^k
Thus 13! is divisible by 24^k if and only if 13! is divisible by 2^(3k) and is divisible by 3^k.
We can tell the greatest power of 2 by which 13! is divisible by if we look at its factors which are divisible by 2:
2 = 2^1
4 = 2^2
6 = 2^1*3
8 = 2^3
10 = 2^1*5
12 = 2^2*3
As none of the odd factors contribute any factors of 2, we have
13! = (2^1*2^2*2^1*2^3*2^1*2^2)*m = 2^(10)*m
where m is some integer not divisible by 2. As such, we know that 13! is divisible by 2^(3k) if and only if 2^10 is divisible by 2^(3k), meaning 3k <= 10. As k is an integer, this means k <= 3.
Next, we can look at which factors of 13! are divisible by 3:
3 = 3^1
6 = 3^1 * 2
9 = 3^2
12 = 3^1*4
As no other factors of 13! contribute any factors of 3, this means
13! = (3^1*3^1*3^2*3^1)*n = 3^5*n
where n is some integer not divisible by 3. As such, we know that 3^5 is divisible by 3^k, meaning k <= 5.
The largest nonnegative integer satisfying the constraints k<=3 and k<=5 is 3, giving us our answer of k=3.
A calculator will verify that (13!)/24^3 = 450450, whereas (13!)/24^4=18768.75
