Find the number of Zeros at the end of the product -
`12xx27xx63xx113xx1250xx24xx650`

Options

A. 2
B. 6
C. 4
D. 8

Answer

To find the number of zeros at the end of the product \( 12 \times 27 \times 63 \times 113 \times 1250 \times 24 \times 650 \), 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. **Factor each number into its prime factors:**
- \( 12 = 2^2 \times 3^1 \)
- \( 27 = 3^3 \)
- \( 63 = 3^2 \times 7^1 \)
- \( 113 = 113^1 \) (prime)
- \( 1250 = 5^4 \times 2^1 \) (since \( 1250 = 125 \times 10 = 5^3 \times (2 \times 5) \))
- \( 24 = 2^3 \times 3^1 \)
- \( 650 = 2^1 \times 5^2 \times 13^1 \) (since \( 650 = 65 \times 10 = (5 \times 13) \times (2 \times 5) \))

2. **Count the total number of factors of 2:**
- From \( 12 \): \( 2^2 \) contributes 2
- From \( 27 \): contributes 0
- From \( 63 \): contributes 0
- From \( 113 \): contributes 0
- From \( 1250 \): \( 2^1 \) contributes 1
- From \( 24 \): \( 2^3 \) contributes 3
- From \( 650 \): \( 2^1 \) contributes 1

Total factors of 2:
\[ 2 + 0 + 0 + 0 + 1 + 3 + 1 = 7 \]

3. **Count the total number of factors of 5:**
- From \( 12 \): contributes 0
- From \( 27 \): contributes 0
- From \( 63 \): contributes 0
- From \( 113 \): contributes 0
- From \( 1250 \): \( 5^4 \) contributes 4
- From \( 24 \): contributes 0
- From \( 650 \): \( 5^2 \) contributes 2

Total factors of 5:
\[ 0 + 0 + 0 + 0 + 4 + 0 + 2 = 6 \]

4. **Determine the number of pairs of (2, 5):**
The number of pairs of 2 and 5 that can be formed is the minimum of the total counts of 2 and 5:
\[ \text{Number of pairs} = \min(7, 6) = 6 \]

5. **Conclusion:**
The number of zeros at the end of the product \( 12 \times 27 \times 63 \times 113 \times 1250 \times 24 \times 650 \) is **6**.