The number of zeros at the end of `1^(99) xx 2^(98) xx 3^(97) xx…99^(1)` is:

Options

A. 1100
B. 1111
C. 1099
D. none of these

Answer

To find the number of zeros at the end of the expression \(1^{99} \times 2^{98} \times 3^{97} \times \ldots \times 99^{1}\), we need to determine how many times the factors of 10 can be formed from the product. Each factor of 10 is made from one factor of 2 and one factor of 5. Therefore, we need to find the minimum of the number of 2s and 5s in the prime factorization of the entire product.

### Step-by-step Solution:

1. **Understanding the Expression**:
The expression can be rewritten as:
\[ P = 1^{99} \times 2^{98} \times 3^{97} \times \ldots \times 99^{1} \]
This means we have each integer from 1 to 99 raised to a power that decreases from 99 to 1.

2. **Finding the Number of 2s**:
We need to find the total number of factors of 2 in the product \(P\). This can be done by counting the contributions from each integer:
\[ \text{Total number of 2s} = \sum_{k=1}^{99} k \cdot v_2(k) \]
where \(v_2(k)\) is the highest power of 2 that divides \(k\).

The contribution of each integer \(k\) to the power of 2 can be calculated as follows:
- For \(k = 1\) to \(99\), we can count how many multiples of \(2, 4, 8, \ldots\) there are.

3. **Finding the Number of 5s**:
Similarly, we need to find the total number of factors of 5 in the product \(P\):
\[ \text{Total number of 5s} = \sum_{k=1}^{99} k \cdot v_5(k) \]
where \(v_5(k)\) is the highest power of 5 that divides \(k\).

The contribution of each integer \(k\) to the power of 5 can be calculated in a similar manner:
- Count how many multiples of \(5, 25, \ldots\) there are.

4. **Calculating the Contributions**:
- For the number of 2s:
- Count multiples of \(2\): \(49\)
- Count multiples of \(4\): \(24\)
- Count multiples of \(8\): \(12\)
- Count multiples of \(16\): \(6\)
- Count multiples of \(32\): \(3\)
- Count multiples of \(64\): \(1\)
- Total = \(49 + 24 + 12 + 6 + 3 + 1 = 95\)

- For the number of 5s:
- Count multiples of \(5\): \(19\)
- Count multiples of \(25\): \(3\)
- Total = \(19 + 3 = 22\)

5. **Finding the Minimum**:
The number of trailing zeros in the product \(P\) is determined by the limiting factor, which is the smaller count between the number of 2s and 5s:
\[ \text{Number of trailing zeros} = \min(95, 22) = 22 \]

### Final Answer:
The number of zeros at the end of \(1^{99} \times 2^{98} \times 3^{97} \times \ldots \times 99^{1}\) is **22**.