Find the number of Zeros at the end of the product -
`140! xx 5 xx 15 xx 22 xx 11 xx 44 xx 135`

Options

A. 34
B. 35
C. 36
D. 37

Answer

To find the number of zeros at the end of the product \( 140! \times 5 \times 15 \times 22 \times 11 \times 44 \times 135 \), we need to determine how many times the factors of 10 can be formed in this product. A factor of 10 is formed by a pair of factors 2 and 5. Since there are generally more factors of 2 than factors of 5 in factorials, we will focus on counting the number of factors of 5.

### Step-by-Step Solution:

1. **Count the factors of 5 in \( 140! \)**:
- To find the number of factors of 5 in \( 140! \), we use the formula:
\[ \text{Number of factors of } 5 = \left\lfloor \frac{140}{5} \right\rfloor + \left\lfloor \frac{140}{25} \right\rfloor + \left\lfloor \frac{140}{125} \right\rfloor \]
- Calculating each term:
- \( \left\lfloor \frac{140}{5} \right\rfloor = \left\lfloor 28 \right\rfloor = 28 \)
- \( \left\lfloor \frac{140}{25} \right\rfloor = \left\lfloor 5.6 \right\rfloor = 5 \)
- \( \left\lfloor \frac{140}{125} \right\rfloor = \left\lfloor 1.12 \right\rfloor = 1 \)
- Adding these values gives:
\[ 28 + 5 + 1 = 34 \]

2. **Count the factors of 5 in the additional numbers**:
- Next, we need to count the factors of 5 in the numbers \( 5, 15, 22, 11, 44, \) and \( 135 \):
- \( 5 \): \( 1 \) factor of 5
- \( 15 \): \( 1 \) factor of 5
- \( 22 \): \( 0 \) factors of 5
- \( 11 \): \( 0 \) factors of 5
- \( 44 \): \( 0 \) factors of 5
- \( 135 \): \( 1 \) factor of 5
- Adding these values gives:
\[ 1 + 1 + 0 + 0 + 0 + 1 = 3 \]

3. **Total factors of 5**:
- Now, we combine the factors of 5 from \( 140! \) and the additional numbers:
\[ 34 + 3 = 37 \]

4. **Conclusion**:
- The total number of zeros at the end of the product \( 140! \times 5 \times 15 \times 22 \times 11 \times 44 \times 135 \) is \( 37 \).

### Final Answer:
The number of zeros at the end of the product is **37**.