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 $n$.