If `n=2^3xx3^4xx5^4xx7`, where n is a natural number, then find the number of consecutive zeros in n

Answer

To find the number of consecutive zeros in the number \( n = 2^3 \times 3^4 \times 5^4 \times 7 \), we need to determine how many times \( n \) can be divided by \( 10 \) without leaving a remainder. Since \( 10 = 2 \times 5 \), we need to find the minimum of the powers of \( 2 \) and \( 5 \) in the prime factorization of \( n \).

### Step-by-step Solution:

1. **Identify the prime factorization of \( n \)**:
\[ n = 2^3 \times 3^4 \times 5^4 \times 7 \]

2. **Count the powers of \( 2 \) and \( 5 \)**:
- The power of \( 2 \) in \( n \) is \( 3 \).
- The power of \( 5 \) in \( n \) is \( 4 \).

3. **Find the minimum of the powers of \( 2 \) and \( 5 \)**:
\[ \text{Number of zeros} = \min(\text{power of } 2, \text{power of } 5) = \min(3, 4) \]

4. **Calculate the minimum**:
\[ \min(3, 4) = 3 \]

5. **Conclusion**:
Therefore, the number of consecutive zeros in \( n \) is \( 3 \).

### Final Answer:
The number of consecutive zeros in \( n \) is \( 3 \).