The number of zeroes at the end of (127)! Is

Options

A. 31
B. 30
C. 0
D. 10

Answer

To find the number of trailing zeros in \(127!\), we need to determine how many times \(10\) is a factor in \(127!\). Since \(10 = 2 \times 5\), and there are always more factors of \(2\) than \(5\) in factorials, we only need to count the number of times \(5\) is a factor in \(127!\).

### Step-by-Step Solution:

1. **Understanding the Formula**:
The number of trailing zeros in \(n!\) can be calculated using the formula:
\[ \text{Number of trailing zeros} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{5^2} \right\rfloor + \left\lfloor \frac{n}{5^3} \right\rfloor + \ldots \]
until \(5^k > n\).

2. **Applying the Formula**:
For \(n = 127\):
- Calculate \(\left\lfloor \frac{127}{5} \right\rfloor\):
\[ \left\lfloor \frac{127}{5} \right\rfloor = \left\lfloor 25.4 \right\rfloor = 25 \]
- Calculate \(\left\lfloor \frac{127}{25} \right\rfloor\):
\[ \left\lfloor \frac{127}{25} \right\rfloor = \left\lfloor 5.08 \right\rfloor = 5 \]
- Calculate \(\left\lfloor \frac{127}{125} \right\rfloor\):
\[ \left\lfloor \frac{127}{125} \right\rfloor = \left\lfloor 1.016 \right\rfloor = 1 \]
- Calculate \(\left\lfloor \frac{127}{625} \right\rfloor\):
\[ \left\lfloor \frac{127}{625} \right\rfloor = \left\lfloor 0.2032 \right\rfloor = 0 \]
Since \(625 > 127\), we stop here.

3. **Summing the Results**:
Now, we sum all the values calculated:
\[ 25 + 5 + 1 + 0 = 31 \]

4. **Conclusion**:
Therefore, the number of trailing zeros in \(127!\) is \(31\).

### Final Answer:
The number of zeros at the end of \(127!\) is **31**.

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