10000! = (100!)^{K}× P, where P and K are integers. What can be the maximum value of K?

Consider,

5! = (3!)^{K}× P

here, k = min [quotient of 3/1, quotient of 1/1] = 1 [there are 3 2s in 5! ; 1 2s in 3! ; 1 3s in 5! ; 1 3s in 3!]

25! = (5!)^{K}× P

here, k = min [quotient of 22/3, quotient of 10/1, quotient of 6/1] = 6 [there are 22 2s in 25! ; 3 2s in 5! ; 10 3s in 5! ; 1 3s in 3!; 6 5s in 25!, 1 5s in 5!]

100! = (10!)^{K}× P

here, k = min [quotient of 97/8, quotient of 48/4, quotient of 24/2, quotient 16/1] = 12

Now returning to the original question,

The maximum value of k will depend on highest or lowest or some other prime containted in 100?

ie will it depend on number of 2s, 97s or some other prime number? and why?

Thank you in advance.