2013 Cayley Problem 25

A positive integer has k *trailing zeros* if its last k digits are all zero and it has a non-zero digit immediately to the left of these k zeros. For example, the number 1\,030\,000 has 4 trailing zeros. Define Z(m) to be the number of trailing zeros of the positive integer m. Lloyd is bored one day, so makes a list of the value of n - Z(n!) for each integer n from 100 to 10\,000, inclusive. How many integers appear in his list at least three times?

(Note: If n is a positive integer, the symbol n! (read "n factorial") is used to represent the product of the integers from 1 to n. That is, n! = n(n-1)(n-2)\cdots(3)(2)(1). For example, 5! = 5(4)(3)(2)(1) or 5! = 120.)

\textbf{(A)}\ 2\quad \textbf{(B)}\ 3\quad \textbf{(C)}\ 4\quad \textbf{(D)}\ 5\quad \textbf{(E)}\ 6

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Cayley contests here.