Zeros of the decimal representation of $k!$ I'd like a hint for the question:

For how many positive integers $k$ does the ordinary decimal representation of the integer $k!$ end in exactly $99$ zeros?

Thanks.
Hint. A number ends in $0$ if and only if it is a multiple of $10$. A number ends in two zeroes if and only if it is a multiple of $100=2^2\times 5^2$. A number ends in three zeroes if and only if it is a multiple of $1000 = 2^3\times 5^3$. Etc. So, e.g.,  a number ends in exactly two zeroes if and only if it is a multiple of $100$ but not of $1000$.
In order that k! end in 99 zeroes, it must have 99 factors of 10 and so 99 2's and 99 5's. Since there are going to be fewer 5's than 2's, k! must have exactly 99 5's. We first get that for k= 99*5= 495 and then get 100 5's for k= 100*5= 500. Any number from 495 to 499 will have exactly 99 5's in its factorial. There are 5 such numbers.

Actually, the answer is correct but the process has a little problem. 25=5^2 provides double 5. 125=5^3 provides triple 5. However,100 and 99 minus 3(1+2 extra 5) at the same time. Hence, it has the same answer with the solution above.