Understanding Consecutive Zeros in Factorials

The number of consecutive zeros at the end of the expansion of a factorial, \(n!\), is determined by the number of times 10 is a factor in its prime factorization. Since \(10 = 2 \times 5\), we need to count the pairs of 2s and 5s. In any factorial \(n!\), the number of factors of 2 is always greater than or equal to the number of factors of 5. Therefore, the number of trailing zeros is simply the count of the factors of 5 in the prime factorization of \(n!\).

Calculating Zeros using Legendre's Formula

We can find the total number of factors of a prime \(p\) in \(n!\) using Legendre's formula:

\(\nu_p(n!) = \sum_{i=1}^{\infty} \left\lfloor \frac{n}{p^i} \right\rfloor\)

For trailing zeros, we are interested in the number of factors of 5, so \(p=5\). The formula becomes:

\(Z(n) = \nu_5(n!) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \dots\)

We are looking for the number of values of the natural number \(n\) for which \(Z(n) = 2\).

Finding Values of n for Exactly 2 Zeros

Let's analyze the formula \(Z(n) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \dots\). We want \(Z(n) = 2\).

Consider the first term, \(\lfloor \frac{n}{5} \rfloor\).

• If \(\lfloor \frac{n}{5} \rfloor = 1\), then \(1 \le \frac{n}{5} < 2\), which means \(5 \le n < 10\). For these values of \(n\), \(\lfloor \frac{n}{25} \rfloor = 0\), \(\lfloor \frac{n}{125} \right\rfloor = 0\), and so on. Thus, \(Z(n) = 1\). This is not what we need.
• If \(\lfloor \frac{n}{5} \rfloor = 2\), then \(2 \le \frac{n}{5} < 3\), which means \(10 \le n < 15\). Let's check the higher terms for this range:
  • For \(n\) in the range \([10, 14]\), the term \(\lfloor \frac{n}{25} \rfloor = 0\) (since \(n < 25\)). Similarly, all subsequent terms are also 0.
  • Therefore, for \(n\) in the range \(10 \le n < 15\), \(Z(n) = \lfloor \frac{n}{5} \rfloor = 2\).
  The values of \(n\) in this range are \(10, 11, 12, 13, 14\).
• If \(\lfloor \frac{n}{5} \rfloor = 3\), then \(3 \le \frac{n}{5} < 4\), which means \(15 \le n < 20\). For these values, \(Z(n) = \lfloor \frac{n}{5} \rfloor = 3\). This is greater than 2.
• If \(\lfloor \frac{n}{5} \rfloor\) is greater than or equal to 3, \(Z(n)\) will be greater than 2.

Now, let's consider the impact of the \(\lfloor \frac{n}{25} \rfloor\) term. The smallest value of \(n\) for which \(\lfloor \frac{n}{25} \rfloor\) is non-zero is \(n=25\).

• For \(n=25\): \(Z(25) = \lfloor \frac{25}{5} \rfloor + \lfloor \frac{25}{25} \rfloor = 5 + 1 = 6\).

Since the number of zeros increases as \(n\) increases, and we already found \(Z(n) = 3\) for \(n \ge 15\), any \(n \ge 25\) will result in \(Z(n) \ge 6\).

Conclusion

The only natural numbers \(n\) for which \(n!\) has exactly 2 consecutive zeros at the end are \(n = 10, 11, 12, 13, 14\). Counting these values, we find there are 5 possible values for \(n\).