Find the number of Zeros at the end of the product -
`1^11 xx 2^21 xx 3^31 xx 4^41 ....... 10^101 `

Options

A. a)51
B. b)10
C. c)5! + 10!
D. d) None of these

Answer

To find the number of zeros at the end of the product \(1^{11} \times 2^{21} \times 3^{31} \times 4^{41} \times \ldots \times 10^{101}\), we need to determine how many times the factors of 10 can be formed from the product. Each factor of 10 is made up of one factor of 2 and one factor of 5. Thus, we need to find the minimum of the total number of factors of 2 and factors of 5 in the product.

### Step-by-Step Solution:

1. **Identify the product**:
The product is \(1^{11} \times 2^{21} \times 3^{31} \times 4^{41} \times 5^{51} \times 6^{61} \times 7^{71} \times 8^{81} \times 9^{91} \times 10^{101}\).

2. **Count the factors of 5**:
- From \(1^{11}\): contributes 0 factors of 5.
- From \(2^{21}\): contributes 0 factors of 5.
- From \(3^{31}\): contributes 0 factors of 5.
- From \(4^{41}\): contributes 0 factors of 5.
- From \(5^{51}\): contributes 51 factors of 5.
- From \(6^{61}\): contributes 0 factors of 5.
- From \(7^{71}\): contributes 0 factors of 5.
- From \(8^{81}\): contributes 0 factors of 5.
- From \(9^{91}\): contributes 0 factors of 5.
- From \(10^{101}\): contributes 101 factors of 5.

**Total factors of 5**: \(0 + 0 + 0 + 0 + 51 + 0 + 0 + 0 + 0 + 101 = 152\).

3. **Count the factors of 2**:
- From \(1^{11}\): contributes 0 factors of 2.
- From \(2^{21}\): contributes 21 factors of 2.
- From \(3^{31}\): contributes 0 factors of 2.
- From \(4^{41}\): \(4 = 2^2\), so contributes \(2 \times 41 = 82\) factors of 2.
- From \(5^{51}\): contributes 0 factors of 2.
- From \(6^{61}\): \(6 = 2 \times 3\), so contributes 61 factors of 2.
- From \(7^{71}\): contributes 0 factors of 2.
- From \(8^{81}\): \(8 = 2^3\), so contributes \(3 \times 81 = 243\) factors of 2.
- From \(9^{91}\): contributes 0 factors of 2.
- From \(10^{101}\): \(10 = 2 \times 5\), so contributes 101 factors of 2.

**Total factors of 2**: \(0 + 21 + 0 + 82 + 0 + 61 + 0 + 243 + 0 + 101 = 508\).

4. **Determine the number of pairs of (2, 5)**:
The number of zeros at the end of the product is determined by the minimum of the total factors of 2 and 5:
\[ \text{Number of zeros} = \min(508, 152) = 152. \]

### Final Answer:
The number of zeros at the end of the product is **152**.