Find the number of Zeros at the end of the product -
`1 xx 3 xx 5 xx 7 xx 9 xx 11 ...... 99xx101 `

Options

A. 24
B. 5
C. 2
D. 0

Answer

To find the number of zeros at the end of the product \(1 \times 3 \times 5 \times 7 \times \ldots \times 99 \times 101\), we need to determine how many times the factors of 10 are present in this product. A factor of 10 is made up of a pair of factors 2 and 5.

### Step-by-step Solution:

1. **Identify the Product**:
The product consists of all odd numbers from 1 to 101. This can be expressed as:
\[ P = 1 \times 3 \times 5 \times 7 \times \ldots \times 99 \times 101 \]

2. **Count the Factors of 2**:
Since all the numbers in this product are odd, there are no even numbers present. Therefore, there are no factors of 2 in this product.
\[ \text{Number of factors of 2} = 0 \]

3. **Count the Factors of 5**:
Next, we need to count the number of factors of 5 in the product. The odd numbers that contribute a factor of 5 are 5, 15, 25, 35, 45, 55, 65, 75, 85, 95, and 105. However, we only consider up to 101.
- The odd multiples of 5 up to 101 are: 5, 15, 25, 35, 45, 55, 65, 75, 85, 95.
- Count these: There are 10 odd multiples of 5.

4. **Determine the Number of Pairs of (2, 5)**:
Since there are no factors of 2 in the product, we cannot form any pairs of (2, 5). Therefore, the number of zeros at the end of the product is determined solely by the number of factors of 2.
\[ \text{Number of zeros} = \min(\text{Number of factors of 2}, \text{Number of factors of 5}) = \min(0, 10) = 0 \]

5. **Conclusion**:
Thus, the number of zeros at the end of the product \(1 \times 3 \times 5 \times \ldots \times 99 \times 101\) is **0**.