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 of these

Answer

To find the number of zeros at the end of the product \( \prod_{n=5}^{30} n^{n+1} \), we need to determine how many times the factor 10 can be formed in the expression. A factor of 10 is produced from the pair of factors 2 and 5. Since there are usually more factors of 2 than factors of 5 in such products, we will focus on counting the number of factors of 5.

### Step-by-Step Solution:

1. **Understand the Product**:
The expression \( \prod_{n=5}^{30} n^{n+1} \) means we are multiplying each integer from 5 to 30 raised to the power of \( n+1 \).

2. **Expand the Product**:
The product can be written as:
\[ 5^6 \cdot 6^7 \cdot 7^8 \cdot 8^9 \cdot 9^{10} \cdot 10^{11} \cdots \cdot 30^{31} \]

3. **Count the Factors of 5**:
We need to count how many times 5 appears as a factor in this product.

- For \( n = 5 \): \( 5^6 \) contributes 6 factors of 5.
- For \( n = 6 \): \( 6^7 \) contributes 0 factors of 5.
- For \( n = 7 \): \( 7^8 \) contributes 0 factors of 5.
- For \( n = 8 \): \( 8^9 \) contributes 0 factors of 5.
- For \( n = 9 \): \( 9^{10} \) contributes 0 factors of 5.
- For \( n = 10 \): \( 10^{11} = (5 \cdot 2)^{11} \) contributes 11 factors of 5.
- For \( n = 11 \): \( 11^{12} \) contributes 0 factors of 5.
- For \( n = 12 \): \( 12^{13} \) contributes 0 factors of 5.
- For \( n = 13 \): \( 13^{14} \) contributes 0 factors of 5.
- For \( n = 14 \): \( 14^{15} \) contributes 0 factors of 5.
- For \( n = 15 \): \( 15^{16} = (5 \cdot 3)^{16} \) contributes 16 factors of 5.
- For \( n = 16 \): \( 16^{17} \) contributes 0 factors of 5.
- For \( n = 17 \): \( 17^{18} \) contributes 0 factors of 5.
- For \( n = 18 \): \( 18^{19} \) contributes 0 factors of 5.
- For \( n = 19 \): \( 19^{20} \) contributes 0 factors of 5.
- For \( n = 20 \): \( 20^{21} = (5 \cdot 4)^{21} \) contributes 21 factors of 5.
- For \( n = 21 \): \( 21^{22} \) contributes 0 factors of 5.
- For \( n = 22 \): \( 22^{23} \) contributes 0 factors of 5.
- For \( n = 23 \): \( 23^{24} \) contributes 0 factors of 5.
- For \( n = 24 \): \( 24^{25} \) contributes 0 factors of 5.
- For \( n = 25 \): \( 25^{26} = (5^2)^{26} \) contributes \( 2 \times 26 = 52 \) factors of 5.
- For \( n = 26 \): \( 26^{27} \) contributes 0 factors of 5.
- For \( n = 27 \): \( 27^{28} \) contributes 0 factors of 5.
- For \( n = 28 \): \( 28^{29} \) contributes 0 factors of 5.
- For \( n = 29 \): \( 29^{30} \) contributes 0 factors of 5.
- For \( n = 30 \): \( 30^{31} = (5 \cdot 6)^{31} \) contributes 31 factors of 5.

4. **Total Count of Factors of 5**:
Now, we sum all the contributions:
\[ 6 + 0 + 0 + 0 + 0 + 11 + 0 + 0 + 0 + 0 + 16 + 0 + 0 + 0 + 0 + 21 + 0 + 0 + 0 + 0 + 52 + 0 + 0 + 0 + 0 + 31 \]
This gives us:
\[ 6 + 11 + 16 + 21 + 52 + 31 = 137 \]

5. **Conclusion**:
The number of zeros at the end of the product \( \prod_{n=5}^{30} n^{n+1} \) is **137**.