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#
