[johnp314]

One can do this mentally easily enough. 52! = 525150....321, and there are 26 even numbers in this progression, so we have 26 powers of 2. Taking them out, there are 13 of those even factors divisible by 4, but we already took out the first 2 from each so we have 13 more 2's, giving 26 + 13 = 39. Now on to factors divisible by 8, they are half of those that are divisible by 4, so half of 13, giving 6 more 2's (52 is not divisible by 8 so we round down). Thus so far we have 39 + 6 = 45 two's in the factorization of 52!. On to numbers less than 52 that are divisible by 16, that's half those divisible by 8, so 3 more, getting us to 48. Finally there is there is the factor 32 = 2^5 of 52! giving us one more 2, hence 49. i.e. for p a prime, the largest k such that p^k divides n! is given by k = Floor(n/p) + Floor(n/p^2) + ... + Floor(n/p^t) where p^(t+1)>n