Number of zeros in the end of `(1^1 xx 2^2 xx 3^3 xx 4^4 xx ....... xx 98^98 xx 99^99 xx 100^100)? `

Options

A. 1200
B. 1300
C. 1500
D. 1600

Answer

To find the number of zeros at the end of the expression \(1^1 \times 2^2 \times 3^3 \times \ldots \times 100^{100}\), we need to determine how many times 10 is a factor in this product. Since \(10 = 2 \times 5\), we need to find the minimum of the number of factors of 2 and the number of factors of 5 in the product.

### Step-by-Step Solution:

1. **Count the number of factors of 5:**
- We will count how many multiples of 5 are present in the numbers from 1 to 100.
- For each multiple of 5, we will also count how many times it contributes to the factor of 5.

The multiples of 5 from 1 to 100 are:
- \(5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100\)

We can count the contributions:
- Each multiple of 5 contributes at least one factor of 5.
- Multiples of \(25\) contribute an additional factor of 5.

Let's count them:
- From \(1\) to \(100\), the multiples of \(5\) are \(5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100\) (20 numbers).
- The multiples of \(25\) are \(25, 50, 75, 100\) (4 numbers).

Therefore, the total number of factors of 5 is:
\[ 20 + 4 = 24 \]

2. **Count the number of factors of 2:**
- Similarly, we will count how many multiples of 2 are present in the numbers from 1 to 100.

The multiples of 2 from 1 to 100 are:
- \(2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100\)

Let's count them:
- Each multiple of \(2\) contributes at least one factor of 2.
- Multiples of \(4\) contribute an additional factor of 2.
- Multiples of \(8\) contribute yet another factor of 2.
- Multiples of \(16\) contribute another factor of 2.
- Multiples of \(32\) contribute another factor of 2.
- Multiples of \(64\) contribute another factor of 2.

Therefore, the total number of factors of 2 is:
\[ 50 + 25 + 12 + 6 + 3 + 1 = 97 \]

3. **Determine the number of trailing zeros:**
- The number of trailing zeros in the product is determined by the limiting factor, which is the smaller of the counts of 2s and 5s.

Thus, the number of trailing zeros is:
\[ \min(24, 97) = 24 \]

### Final Answer:
The number of zeros at the end of \(1^1 \times 2^2 \times 3^3 \times \ldots \times 100^{100}\) is **24**.