Find the number of zeroes present at the end on 100!

Answer

To find the number of zeros at the end of \(100!\), we need to determine how many times \(10\) can be factored from \(100!\). Since \(10 = 2 \times 5\), we need to count the pairs of factors \(2\) and \(5\) in \(100!\). However, there will always be more factors of \(2\) than \(5\) in factorials, so we only need to count the number of times \(5\) appears as a factor.

### Step-by-Step Solution:

1. **Identify the formula**: To find the number of times a prime \(p\) divides \(n!\), we use the formula:
\[ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]
where \(n\) is the number we are taking the factorial of (in this case, \(100\)) and \(p\) is the prime factor we are interested in (in this case, \(5\)).

2. **Calculate for \(p = 5\)**:
- For \(k = 1\):
\[ \left\lfloor \frac{100}{5^1} \right\rfloor = \left\lfloor \frac{100}{5} \right\rfloor = \left\lfloor 20 \right\rfloor = 20 \]
- For \(k = 2\):
\[ \left\lfloor \frac{100}{5^2} \right\rfloor = \left\lfloor \frac{100}{25} \right\rfloor = \left\lfloor 4 \right\rfloor = 4 \]
- For \(k = 3\):
\[ \left\lfloor \frac{100}{5^3} \right\rfloor = \left\lfloor \frac{100}{125} \right\rfloor = \left\lfloor 0.8 \right\rfloor = 0 \]
- Since \(5^3\) exceeds \(100\), we can stop here.

3. **Sum the results**:
\[ 20 + 4 + 0 = 24 \]

4. **Conclusion**: The number of zeros at the end of \(100!\) is \(24\).

### Final Answer:
The number of zeros present at the end of \(100!\) is \(24\).