How many zeroes are there in the product` 48 xx125xx125xx625xx32xx128`?

Options

A. a)10
B. b)12
C. c)14
D. d)20

Answer

To find the number of zeroes in the product \( 48 \times 125 \times 125 \times 625 \times 32 \times 128 \), we need to determine how many pairs of the factors 2 and 5 can be formed, as each pair contributes to one trailing zero in the product.

### Step-by-Step Solution:

1. **Prime Factorization of Each Number**:
- **48**:
\[ 48 = 16 \times 3 = 2^4 \times 3^1 \]
- **125**:
\[ 125 = 5^3 \]
- **125** (again):
\[ 125 = 5^3 \]
- **625**:
\[ 625 = 25 \times 25 = 5^4 \]
- **32**:
\[ 32 = 2^5 \]
- **128**:
\[ 128 = 2^7 \]

2. **Combine the Prime Factors**:
- Now we will sum the powers of 2 and 5 from all the numbers:
- Total power of \(2\):
\[ 4 \text{ (from 48)} + 5 \text{ (from 32)} + 7 \text{ (from 128)} = 16 \]
- Total power of \(5\):
\[ 3 \text{ (from first 125)} + 3 \text{ (from second 125)} + 4 \text{ (from 625)} = 10 \]

3. **Finding the Number of Pairs**:
- The number of pairs of \(2\) and \(5\) that can be formed is determined by the smaller of the two totals:
\[ \text{Number of pairs} = \min(16, 10) = 10 \]

4. **Conclusion**:
- Therefore, the number of trailing zeroes in the product \( 48 \times 125 \times 125 \times 625 \times 32 \times 128 \) is **10**.