Let `p(n) = (n+1)(n+3) (n+5)(n+7)(n+9)` . What is the largest integer that is a divisor of p(n) for all positive even integers n ?

Answer

To find the largest integer that is a divisor of \( p(n) = (n+1)(n+3)(n+5)(n+7)(n+9) \) for all positive even integers \( n \), we can follow these steps:

### Step 1: Substitute \( n \) with an even integer
Let \( n = 2k \), where \( k \) is a positive integer. Then we can rewrite \( p(n) \) as follows:
\[ p(2k) = (2k + 1)(2k + 3)(2k + 5)(2k + 7)(2k + 9) \]

### Step 2: Analyze the product
The expression \( p(2k) \) consists of five consecutive odd numbers:
- \( 2k + 1 \)
- \( 2k + 3 \)
- \( 2k + 5 \)
- \( 2k + 7 \)
- \( 2k + 9 \)

### Step 3: Determine the divisibility by 3
Among any five consecutive odd numbers, at least one of them must be divisible by 3. Therefore, \( p(2k) \) is divisible by 3.

### Step 4: Determine the divisibility by 5
Similarly, among any five consecutive odd numbers, at least one of them must also be divisible by 5. Therefore, \( p(2k) \) is also divisible by 5.

### Step 5: Combine the results
Since \( p(2k) \) is divisible by both 3 and 5, it follows that \( p(2k) \) is divisible by:
\[ 3 \times 5 = 15 \]

### Step 6: Check for higher divisibility
Next, we check if there are any higher divisors that \( p(n) \) might consistently have for all positive even integers \( n \). The product of five consecutive odd numbers can be analyzed for higher powers of 3 and 5, but since we are looking for the largest integer that divides \( p(n) \) for all positive even integers \( n \), we find that no higher common factor consistently divides all such products.

### Conclusion
Thus, the largest integer that is a divisor of \( p(n) \) for all positive even integers \( n \) is:
\[ \boxed{15} \]