If (n-5) is divisible by 17 for every n`in``I^+` then the greatest integer which will necessarily divide (n+12)(n+29) is :

Options

A. 578
B. 289
C. such a number does not exist
D. none of the above

Answer

To solve the problem, we need to find the greatest integer that will necessarily divide the expression \((n+12)(n+29)\) given that \((n-5)\) is divisible by 17 for every positive integer \(n\).

### Step-by-Step Solution:

1. **Understanding the Condition**:
We know that \((n-5)\) is divisible by 17. This means we can express \(n\) in terms of 17:
\[ n = 17k + 5 \]
where \(k\) is any integer.

2. **Substituting \(n\) into the Expression**:
We need to evaluate \((n+12)(n+29)\):
\[ n + 12 = (17k + 5) + 12 = 17k + 17 = 17(k + 1) \]
\[ n + 29 = (17k + 5) + 29 = 17k + 34 = 17(k + 2) \]

3. **Forming the Product**:
Now substitute these into the product:
\[ (n + 12)(n + 29) = (17(k + 1))(17(k + 2)) = 17^2(k + 1)(k + 2) \]

4. **Finding the Greatest Integer**:
The expression \(17^2(k + 1)(k + 2)\) indicates that the product is divisible by \(17^2\). Since \(k + 1\) and \(k + 2\) are integers, they do not affect the divisibility by \(17^2\).

5. **Conclusion**:
Therefore, the greatest integer that will necessarily divide \((n + 12)(n + 29)\) for every positive integer \(n\) such that \((n - 5)\) is divisible by 17 is:
\[ \boxed{289} \]