The number of zeros at the end of 100! Is :

Options

A. 25
B. 50
C. 24
D. 100

Answer

To find the number of zeros at the end of \(100!\) (100 factorial), we need to determine how many times \(10\) is a factor in the product of all integers from \(1\) to \(100\). Since \(10\) is made up of \(2\) and \(5\), and there are generally more factors of \(2\) than \(5\) in factorials, we only need to count the number of times \(5\) is a factor in \(100!\).

### Step-by-Step Solution:

1. **Understanding Factorial**:
\[ 100! = 100 \times 99 \times 98 \times \ldots \times 1 \]
We need to find how many times \(10\) can be formed, which is the product of \(2\) and \(5\).

2. **Counting Factors of 5**:
To find the number of \(5\)s in \(100!\), we use the formula:
\[ \text{Number of factors of } 5 = \left\lfloor \frac{100}{5} \right\rfloor + \left\lfloor \frac{100}{25} \right\rfloor + \left\lfloor \frac{100}{125} \right\rfloor \]

3. **Calculating Each Term**:
- First term:
\[ \left\lfloor \frac{100}{5} \right\rfloor = \left\lfloor 20 \right\rfloor = 20 \]
- Second term:
\[ \left\lfloor \frac{100}{25} \right\rfloor = \left\lfloor 4 \right\rfloor = 4 \]
- Third term:
\[ \left\lfloor \frac{100}{125} \right\rfloor = \left\lfloor 0.8 \right\rfloor = 0 \]

4. **Summing the Factors**:
Now, we add these results together:
\[ 20 + 4 + 0 = 24 \]

5. **Conclusion**:
Therefore, the number of zeros at the end of \(100!\) is \(24\).

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

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