Number of zero's at the ends of ` prod _(n=5)^(30)(n)^(n+1) ` is :

Options

A. 111
B. 147
C. 137
D. None

Answer

To find the number of zeros at the end of the expression \( \prod_{n=5}^{30} n^{n+1} \), we need to determine how many times 10 is a factor in the product. Since \( 10 = 2 \times 5 \), we will find the minimum of the number of factors of 2 and 5 in the product, as the limiting factor will determine the number of trailing zeros.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression can be rewritten as:
\[ \prod_{n=5}^{30} n^{n+1} = 5^6 \cdot 6^7 \cdot 7^8 \cdots \cdot 30^{31} \]

2. **Finding Factors of 5**:
We will count how many times 5 appears as a factor in the product. We will consider each term \( n^{n+1} \) for \( n = 5 \) to \( 30 \).

- For \( n = 5 \): \( 5^6 \) contributes 6 factors of 5.
- For \( n = 10 \): \( 10^{11} = (5 \cdot 2)^{11} \) contributes 11 factors of 5.
- For \( n = 15 \): \( 15^{16} = (3 \cdot 5)^{16} \) contributes 16 factors of 5.
- For \( n = 20 \): \( 20^{21} = (4 \cdot 5)^{21} \) contributes 21 factors of 5.
- For \( n = 25 \): \( 25^{26} = (5^2)^{26} = 5^{52} \) contributes 52 factors of 5.
- For \( n = 30 \): \( 30^{31} = (6 \cdot 5)^{31} \) contributes 31 factors of 5.

Now, we sum these contributions:
\[ 6 + 11 + 16 + 21 + 52 + 31 = 137 \]

3. **Finding Factors of 2**:
Next, we will count how many times 2 appears as a factor in the product. However, since factors of 2 will be more abundant than factors of 5, we can skip the detailed counting for this problem.

4. **Conclusion**:
The number of trailing zeros in the product is determined by the number of factors of 5, which we calculated to be 137.

Thus, the number of zeros at the end of \( \prod_{n=5}^{30} n^{n+1} \) is **137**.