The number of zeros which are not possible at the end of the n! is:

Options

A. A. 82
B. B. 73
C. C. 156
D. D. none

Answer

To find the number of zeros that are not possible at the end of \( n! \), we need to understand how zeros are formed in factorials. Zeros at the end of a number are produced by the factors of 10 in that number, and since \( 10 = 2 \times 5 \), we need to count the pairs of 2s and 5s in the prime factorization of \( n! \).

### Step-by-Step Solution:

1. **Understanding Factorials and Zeros**:
- A factorial \( n! \) is the product of all positive integers up to \( n \).
- To form a zero, we need a pair of factors \( 2 \) and \( 5 \).

2. **Counting Factors of 5**:
- The number of zeros at the end of \( n! \) is determined by the number of pairs of \( 2 \) and \( 5 \).
- Since there are generally more factors of \( 2 \) than \( 5 \) in \( n! \), the limiting factor is the number of \( 5s \).
- We can count the number of \( 5s \) in \( n! \) using the formula:
\[ \text{Count of 5s} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \ldots \]
- This counts how many multiples of \( 5, 25, 125, \ldots \) are present in the numbers from \( 1 \) to \( n \).

3. **Counting Factors of 2**:
- Similarly, we can count the number of \( 2s \) in \( n! \) using:
\[ \text{Count of 2s} = \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{4} \right\rfloor + \left\lfloor \frac{n}{8} \right\rfloor + \ldots \]

4. **Calculating Zeros in \( n! \)**:
- The number of trailing zeros \( Z(n) \) in \( n! \) is given by:
\[ Z(n) = \min(\text{Count of 2s}, \text{Count of 5s}) \]
- Since we know that the count of \( 2s \) will always be greater than or equal to the count of \( 5s \), we can simplify this to:
\[ Z(n) = \text{Count of 5s} \]

5. **Finding Zeros Not Possible**:
- To find the number of zeros that are not possible at the end of \( n! \), we need to consider the next integer \( n+1 \) and calculate \( Z(n+1) \).
- The difference \( Z(n+1) - Z(n) \) gives us the number of zeros that are not possible at the end of \( n! \).

### Example Calculation:
Let's take \( n = 299 \) and \( n = 300 \) as per the video transcript.

- For \( n = 299 \):
\[ \text{Count of 5s} = \left\lfloor \frac{299}{5} \right\rfloor + \left\lfloor \frac{299}{25} \right\rfloor + \left\lfloor \frac{299}{125} \right\rfloor = 59 + 11 + 2 = 72 \]

- For \( n = 300 \):
\[ \text{Count of 5s} = \left\lfloor \frac{300}{5} \right\rfloor + \left\lfloor \frac{300}{25} \right\rfloor + \left\lfloor \frac{300}{125} \right\rfloor = 60 + 12 + 2 = 74 \]

- Thus, the number of zeros not possible at the end of \( 299! \) is:
\[ Z(300) - Z(299) = 74 - 72 = 2 \]

### Conclusion:
The number of zeros that are not possible at the end of \( n! \) for \( n = 299 \) is 2.