How many zeroes are there in 80! ?

Options

A. 16
B. 18
C. 19
D. 20

Answer

To find the number of trailing zeros in \(80!\), we need to determine how many times \(10\) is a factor in the numbers from \(1\) to \(80\). Since \(10\) is made up of \(2 \times 5\), 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 \(80!\).

### Step-by-Step Solution:

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

2. **Calculate for \(n = 80\)**:
- First, calculate \(\left\lfloor \frac{80}{5} \right\rfloor\):
\[ \left\lfloor \frac{80}{5} \right\rfloor = \left\lfloor 16 \right\rfloor = 16 \]
- Next, calculate \(\left\lfloor \frac{80}{25} \right\rfloor\):
\[ \left\lfloor \frac{80}{25} \right\rfloor = \left\lfloor 3.2 \right\rfloor = 3 \]
- Finally, calculate \(\left\lfloor \frac{80}{125} \right\rfloor\):
\[ \left\lfloor \frac{80}{125} \right\rfloor = \left\lfloor 0.64 \right\rfloor = 0 \]

3. **Sum the results**:
Now, add the results from the calculations:
\[ 16 + 3 + 0 = 19 \]

4. **Conclusion**:
Therefore, the number of trailing zeros in \(80!\) is \(19\).

### Final Answer:
The number of trailing zeros in \(80!\) is **19**.