For how many positive integral values of n does n! end with precisely 25 zeros?

Explanation

To determine how many positive integral values of \( n \) result in \( n! \) ending with exactly 25 zeros, we need to understand how trailing zeros are formed in factorials. Trailing zeros in \( n! \) are produced by the factors of 10, which are made up of pairs of factors 2 and 5. Since there are usually more factors of 2 than factors of 5 in factorials, the number of trailing zeros is determined by the number of times 5 is a factor in the numbers from 1 to \( n \).

### Step-by-Step Solution:

1. **Formula for Trailing Zeros**:
The number of trailing zeros \( Z(n) \) in \( n! \) can be calculated using the formula:
\[
Z(n) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \ldots
...