Find the maximum value of n such that
`570xx60xx30xx90xx100xx50xx700xx343xx720xx81` is perfectly divisible by `30^(n)`.

Options

A. 12
B. 11
C. 14
D. 13

Answer

To find the maximum value of \( n \) such that the product \( 570 \times 60 \times 30 \times 90 \times 100 \times 50 \times 700 \times 343 \times 720 \times 81 \) is perfectly divisible by \( 30^n \), we first need to factor \( 30 \) into its prime factors:

\[ 30 = 2 \times 3 \times 5 \]

Thus, \( 30^n = 2^n \times 3^n \times 5^n \). To determine the maximum \( n \), we need to find the minimum power of \( 2 \), \( 3 \), and \( 5 \) in the prime factorization of the product.

### Step 1: Factor each number in the product

1. **570**:
\[ 570 = 2 \times 3 \times 5 \times 19 \]
(Powers: \( 2^1, 3^1, 5^1 \))

2. **60**:
\[ 60 = 2^2 \times 3^1 \times 5^1 \]
(Powers: \( 2^2, 3^1, 5^1 \))

3. **30**:
\[ 30 = 2^1 \times 3^1 \times 5^1 \]
(Powers: \( 2^1, 3^1, 5^1 \))

4. **90**:
\[ 90 = 2^1 \times 3^2 \times 5^1 \]
(Powers: \( 2^1, 3^2, 5^1 \))

5. **100**:
\[ 100 = 2^2 \times 5^2 \]
(Powers: \( 2^2, 5^2 \))

6. **50**:
\[ 50 = 2^1 \times 5^2 \]
(Powers: \( 2^1, 5^2 \))

7. **700**:
\[ 700 = 2^1 \times 5^2 \times 7 \]
(Powers: \( 2^1, 5^2 \))

8. **343**:
\[ 343 = 7^3 \]
(No contribution to \( 2, 3, 5 \))

9. **720**:
\[ 720 = 2^4 \times 3^2 \times 5^1 \]
(Powers: \( 2^4, 3^2, 5^1 \))

10. **81**:
\[ 81 = 3^4 \]
(Powers: \( 3^4 \))

### Step 2: Sum the powers of each prime factor

Now, we sum the powers of \( 2 \), \( 3 \), and \( 5 \):

- **Power of \( 2 \)**:
\[ 1 + 2 + 1 + 1 + 2 + 1 + 1 + 0 + 4 + 0 = 13 \]

- **Power of \( 3 \)**:
\[ 1 + 1 + 1 + 2 + 0 + 0 + 0 + 0 + 2 + 4 = 11 \]

- **Power of \( 5 \)**:
\[ 1 + 1 + 1 + 1 + 2 + 2 + 2 + 0 + 1 + 0 = 11 \]

### Step 3: Determine the minimum power

Now, we find the minimum of these sums to determine \( n \):

\[ n = \min(13, 11, 11) = 11 \]

Thus, the maximum value of \( n \) such that the product is perfectly divisible by \( 30^n \) is:

\[ \boxed{11} \]