Find the number of Zeros at the end of the product -
`20 xx 40 xx 7600 xx 600 xx 300 xx 1000 `

Options

A. 11
B. 10
C. 2
D. 3

Answer

To find the number of zeros at the end of the product \(20 \times 40 \times 7600 \times 600 \times 300 \times 1000\), we need to determine how many times the product can be divided by 10. Each 10 is made up of a factor of 2 and a factor of 5. Therefore, the number of zeros at the end of the product is determined by the minimum of the number of 2s and 5s in the prime factorization of the product.

### Step-by-Step Solution:

1. **Prime Factorization of Each Number:**
- \(20 = 2^2 \times 5^1\)
- \(40 = 2^3 \times 5^1\)
- \(7600 = 2^4 \times 5^2 \times 19^1\)
- \(600 = 2^3 \times 5^2 \times 3^1\)
- \(300 = 2^2 \times 5^2 \times 3^1\)
- \(1000 = 2^3 \times 5^3\)

2. **Count the Total Number of Factors of 2:**
- From \(20\): \(2^2\) contributes 2
- From \(40\): \(2^3\) contributes 3
- From \(7600\): \(2^4\) contributes 4
- From \(600\): \(2^3\) contributes 3
- From \(300\): \(2^2\) contributes 2
- From \(1000\): \(2^3\) contributes 3

Total factors of 2 = \(2 + 3 + 4 + 3 + 2 + 3 = 17\)

3. **Count the Total Number of Factors of 5:**
- From \(20\): \(5^1\) contributes 1
- From \(40\): \(5^1\) contributes 1
- From \(7600\): \(5^2\) contributes 2
- From \(600\): \(5^2\) contributes 2
- From \(300\): \(5^2\) contributes 2
- From \(1000\): \(5^3\) contributes 3

Total factors of 5 = \(1 + 1 + 2 + 2 + 2 + 3 = 11\)

4. **Determine the Number of Zeros:**
The number of zeros at the end of the product is the minimum of the total factors of 2 and 5.
\[ \text{Number of zeros} = \min(17, 11) = 11 \]

### Final Answer:
The number of zeros at the end of the product \(20 \times 40 \times 7600 \times 600 \times 300 \times 1000\) is **11**.