A complete explanation is probably in order here:
Any time (2 * 5) appears in the prime factorization of a number, that number has a "trailing zero".
For example, consider the number 30. 30 = 3 * 2 * 5 = 3 * (2 * 5) = 3 * 10 = 30.
Now consider 300. 300 = 3 * 2 * 5 * 2 * 5 = 3 * (2 * 5) * (2 * 5) = 3 * 10 * 10 = 300.
Now consider 10!
10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
but we'll write it as
9 * 8 * 7 * 6 * 4 * 3 * 1 *
10 * 2 * 5
So 10! has two trailing zeros (!
)
As you've probably gathered, any factorial is going to have more factors of 2 than it does of 5, so for every 5 we find we can easily find a 2 to pair with it.
Hence the question is really asking "How many 5's are there in the prime factorization of (20!*21!*22!*...*32!*33!)�?"
Now notice another shortcut. Every factorial here is a multiple of 20! (For instance, 21! = 21 * 20!.) So 20!, which has four factors of 15 (in its factors of 20, 15, 10, and 5, respectively) will have the same number of 5's as 21!, 22!, 23!, and 24!
So from 20! to 24!, we have 5 * 4 = 20 factors of 5.
25!, however, has two more 5's (since 25 = 5*5). 26! through 29! each have the same number of 5's in their factorizations, so we have another 5 * 6 = 30 factors of 5.
30! has another 5 in the 30, giving it seven factors of 5. 31!, 32!, and 33! are the same, so this gives us another 7 * 4 or 28 factors of 5.
So (20! * ... * 33!) has 20+30+28 = 78 factors of 5.
Since we're raising this to the 6th, we have 78*6 = 468 factors of 5, so we have 468 trailing zeros.
