The factorial function, denoted \( n! \), where \( n \in \mathbb{Z}^* \), is defined as follows: \[ n! = \begin{cases} 1 & \text{if $n = 0$,} \\ n(n - 1)! & \text{otherwise}. \end{cases} \] In the case of \( n = 10 \), \( 10! = 3\,628\,800 \). Note that that value has two trailing zeros. When \( n = 20 \), \( 20! = 2\,432\,902\,008\,176\,640\,000 \) and has four trailing zeros. Can we devise an algorithm to determine how many trailing zeroes there are in \( n! \) without calculating \( n! \)?

Let us define a function \( z : \mathbb{N} \to \mathbb{N} \) that accepts an argument \( n \) that yields the number of trailing zeros in \( n! \). So, using our two examples above, we know that \( z(10) = 2 \) and \( z(20) = 4 \). Now, what is \( z(100) \)?

An integer \( a \) has \( k \) trailing zeros iff \( a = b \cdot 10^k \) for some integer \( b \). Now, the prime factors of \( 10 \) are \( 2 \) and \( 5 \). So, what if we consider the prime factors of each factor in a factorial product? For example, consider the first six terms (after \( 1 \)) of \( 100! \) and their prime factors. We omit \( 1 \) because of its idempotence: \[ \begin{align} 2 &= 2 \\ 3 &= 3 \\ 4 &= 2^2 \\ 5 &= 5 \\ 6 &= 2 \cdot 3 \\ 7 &= 7 \end{align} \] We see that \( 7! \) contains one \( 5 \) and (more than) one \( 2 \). Therefore, \( 7! \) contains a factor of \( 10 \), so we should expect there to be one trailing zero in \( 7! \). And we see that \( 7! = 5040 \), which does indeed have one trailing zero. This suggests that we could, for any integer \( n \) in \( n! \), cycle through the integers from \( 2 \) to \( n \), finding the prime factorization of each integer, and counting pairs of \( 2 \) and \( 5 \) that we find.

You might have noticed from our example of \( 7! \) that there are four \( 2 \)s and only one \( 5 \). Consider that there are \( n/2 \) even factors in \( n! \)—that is, every other integer factor of \( n! \), starting at \(2\), is even. So there are \( n/2 \) even factors. Each of these has at least one \( 2 \) in its prime factorization, and many of them have more than one. Now consider how many factors of \( 5 \) there are in \( n! \): there are, in fact, \( n / 5 \). So, we can see that, for every \( 5 \) we find in \( n! \), there is a \( 2 \) we can match with it. This suggests that we only need count the \( 5 \)s and not the pairs of \( 2 \) and \( 5 \).

This now suggests an algorithm, presented here in Java:

Using this algorithm, we find that \( z(100) = 24 \).

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