Find the number of Zeros at the end of the given expression?
`a = 1^3, b = 2^4, c = 3^5, ...... , z = 26^(28) , a xx b xx c xx d ...... xx z`

Options

A. 112
B. 110
C. 118
D. 115

Answer

To find the number of zeros at the end of the expression \( a \times b \times c \times \ldots \times z \), where \( a = 1^3, b = 2^4, c = 3^5, \ldots, z = 26^{28} \), we need to determine how many times the expression can be divided by 10. Since \( 10 = 2 \times 5 \), we will count the number of factors of 2 and 5 in the product and take the minimum of these two counts.

### Step-by-Step Solution:

1. **Identify the expression**:
The expression is \( a \times b \times c \times \ldots \times z = 1^3 \times 2^4 \times 3^5 \times \ldots \times 26^{28} \).

2. **General term**:
The general term for the product can be expressed as \( n^{(n+2)} \) for \( n = 1, 2, \ldots, 26 \).

3. **Count the number of factors of 2**:
We will find the total power of 2 in the product:
\[ \text{Power of 2} = \sum_{n=1}^{26} (n+2) \cdot \text{(number of factors of 2 in } n) \]
- Calculate the number of factors of 2 for each \( n \) from 1 to 26 and multiply by \( n+2 \).

4. **Count the number of factors of 5**:
Similarly, we will find the total power of 5 in the product:
\[ \text{Power of 5} = \sum_{n=1}^{26} (n+2) \cdot \text{(number of factors of 5 in } n) \]
- Calculate the number of factors of 5 for each \( n \) from 1 to 26 and multiply by \( n+2 \).

5. **Calculate the minimum of the two counts**:
The number of trailing zeros in the product is given by:
\[ \text{Number of trailing zeros} = \min(\text{Power of 2}, \text{Power of 5}) \]

### Detailed Calculation:

- **Power of 2 Calculation**:
- For \( n = 1 \): \( 1^3 \) contributes \( 0 \)
- For \( n = 2 \): \( 2^4 \) contributes \( 4 \)
- For \( n = 3 \): \( 3^5 \) contributes \( 0 \)
- For \( n = 4 \): \( 4^6 \) contributes \( 12 \) (since \( 4 = 2^2 \))
- Continue this for \( n = 1 \) to \( 26 \).

- **Power of 5 Calculation**:
- For \( n = 1 \): \( 1^3 \) contributes \( 0 \)
- For \( n = 2 \): \( 2^4 \) contributes \( 0 \)
- For \( n = 3 \): \( 3^5 \) contributes \( 0 \)
- For \( n = 4 \): \( 4^6 \) contributes \( 0 \)
- For \( n = 5 \): \( 5^7 \) contributes \( 7 \)
- Continue this for \( n = 1 \) to \( 26 \).

6. **Final Calculation**:
- After calculating the contributions for all \( n \), sum them up for both powers.
- Compare the total powers of 2 and 5 to find the minimum.

### Conclusion:
The number of trailing zeros in the product \( a \times b \times c \times \ldots \times z \) is the minimum of the total powers of 2 and 5 calculated.