The number of zeros at the end of 70!, is

Options

A. 16
B. 15
C. 14
D. 20

Answer

To find the number of zeros at the end of \(70!\), we need to determine how many times \(10\) is a factor in \(70!\). 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 \(70!\).

### Step-by-Step Solution:

1. **Identify the formula**: The number of times a prime \(p\) divides \(n!\) can be calculated using the formula:
\[ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]
Here, \(n = 70\) and \(p = 5\).

2. **Calculate the first term**:
\[ \left\lfloor \frac{70}{5} \right\rfloor = \left\lfloor 14 \right\rfloor = 14 \]

3. **Calculate the second term**:
\[ \left\lfloor \frac{70}{5^2} \right\rfloor = \left\lfloor \frac{70}{25} \right\rfloor = \left\lfloor 2.8 \right\rfloor = 2 \]

4. **Calculate the third term**:
\[ \left\lfloor \frac{70}{5^3} \right\rfloor = \left\lfloor \frac{70}{125} \right\rfloor = \left\lfloor 0.56 \right\rfloor = 0 \]
Since \(5^3 = 125\) is greater than \(70\), we stop here.

5. **Sum the results**: Now, we add the results from the terms we calculated:
\[ 14 + 2 + 0 = 16 \]

Thus, the number of zeros at the end of \(70!\) is **16**.

### Final Answer:
The number of zeros at the end of \(70!\) is **16**.

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