If the prime factorization of a natural number `n` is `2^3*3^2*5^2*6,` write the number of consecutive zeros in `ndot`

Explanation

To solve the problem, we need to find the number of consecutive zeros in the natural number \( n \) given its prime factorization. The prime factorization of \( n \) is \( 2^3 \times 3^2 \times 5^2 \times 6 \).

### Step-by-Step Solution:

1. **Understanding the Prime Factorization**:
The prime factorization of \( n \) is given as:
\[
n = 2^3 \times 3^2 \times 5^2 \times 6
...