Find the number of Zeros at the end of the product -
25!x32!x45!`

Options

A. 10
B. 23
C. 22
D. 7

Answer

To find the number of zeros at the end of the product \( 25! \times 32! \times 45! \), we need to determine how many times the factors of 10 can be formed in this product. A factor of 10 is produced by pairing a factor of 2 with a factor of 5. Since there are generally more factors of 2 than factors of 5 in factorials, the number of zeros will be determined by the number of times 5 is a factor in the factorials.

### Step-by-Step Solution:

1. **Calculate the number of factors of 5 in \( 25! \)**:
- The number of factors of 5 in \( n! \) can be found using the formula:
\[ \text{Number of factors of 5} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \ldots \]
- For \( 25! \):
\[ \left\lfloor \frac{25}{5} \right\rfloor + \left\lfloor \frac{25}{25} \right\rfloor = 5 + 1 = 6 \]

2. **Calculate the number of factors of 5 in \( 32! \)**:
- For \( 32! \):
\[ \left\lfloor \frac{32}{5} \right\rfloor + \left\lfloor \frac{32}{25} \right\rfloor = 6 + 1 = 7 \]

3. **Calculate the number of factors of 5 in \( 45! \)**:
- For \( 45! \):
\[ \left\lfloor \frac{45}{5} \right\rfloor + \left\lfloor \frac{45}{25} \right\rfloor = 9 + 1 = 10 \]

4. **Sum the number of factors of 5 from all three factorials**:
- Total number of factors of 5:
\[ 6 \, (\text{from } 25!) + 7 \, (\text{from } 32!) + 10 \, (\text{from } 45!) = 23 \]

5. **Conclusion**:
- The total number of zeros at the end of the product \( 25! \times 32! \times 45! \) is **23**.

### Final Answer:
The number of zeros at the end of the product \( 25! \times 32! \times 45! \) is **23**.