The numbers 2, 4, 6, 8 ...., 98. 100 are multiplied together. The number of zeros at the end of the product must be :

Options

A. 13
B. 12
C. 11
D. 10

Answer

To find the number of zeros at the end of the product of the numbers 2, 4, 6, ..., 100, 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. Since the sequence consists of even numbers, there will be plenty of factors of 2. Therefore, we need to focus on counting the number of factors of 5 in the product.

### Step-by-Step Solution:

1. **Identify the sequence of numbers**: The numbers we are multiplying are 2, 4, 6, 8, ..., 100. This is an arithmetic sequence of even numbers from 2 to 100.

2. **Count the total numbers in the sequence**: The sequence can be expressed as \(2n\) where \(n\) ranges from 1 to 50. Therefore, there are 50 terms in this sequence.

3. **Count the factors of 5**: To find the number of zeros at the end of the product, we need to count how many times 5 is a factor in these numbers.

- The multiples of 5 in the sequence are: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.
- There are 10 multiples of 5.

4. **Count the multiples of 25**: Each multiple of 25 contributes an additional factor of 5. The multiples of 25 in the sequence are: 50, 100.
- There are 2 multiples of 25.

5. **Total count of factors of 5**:
- From multiples of 5: 10
- From multiples of 25: 2
- Total factors of 5 = 10 + 2 = 12.

6. **Conclusion**: Since we have more than enough factors of 2, the number of zeros at the end of the product is determined by the number of factors of 5, which is 12.

### Final Answer:
The number of zeros at the end of the product is **12**.