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

Options

A. The number of 6-digit numbers formed using all the digits 1, 2, 3, 4, 5, 6 divided by `""^(10)C_(2)`
B. The number of 6- digit numbers formed using all the digits 1, 2, 3, 4, 5, 6 divided by 10.
C. Twice the number of 4-digit numbers formed using 4 of the digits 1, 2, 3, 4, 5, 6 divided by `""^(10)C_(2)`
D. The number of 4- digit numbers formed using 4 of the digits 1, 2, 3, 4, 5, 6 divided by 5

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 the product of \(2\) and \(5\), we need to find the number of pairs of \(2\) and \(5\) in the prime factorization of \(70!\). However, there are always more \(2\)s than \(5\)s in factorials, so we only need to count the number of \(5\)s.

### Step-by-Step Solution:

1. **Count the number of factors of \(5\) in \(70!\)**:
We use the formula:
\[ \text{Number of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]
where \(p\) is a prime number.

Here, \(n = 70\) and \(p = 5\).

2. **Calculate each term**:
- First term:
\[ \left\lfloor \frac{70}{5} \right\rfloor = \left\lfloor 14 \right\rfloor = 14 \]
- Second term:
\[ \left\lfloor \frac{70}{25} \right\rfloor = \left\lfloor 2.8 \right\rfloor = 2 \]
- Third term:
\[ \left\lfloor \frac{70}{125} \right\rfloor = \left\lfloor 0.56 \right\rfloor = 0 \]

3. **Add the results**:
Now, we add the results from each term:
\[ 14 + 2 + 0 = 16 \]

4. **Conclusion**:
Therefore, 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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