Question

What is the largest positive integer n such that \(\frac{n^2+7n+12}{n^2-n-12}\)  is also a positive integer?

• A. 8
• B. 12
• C. 16
• D. 6

Answer 1

The Correct Option is B

\(\frac{n^2+7n+12}{n^2-n-12} = \frac{(n+3)(n+4)}{(n-4)(n+3)} \)

\(=\frac{n+4}{n-4} \)

\(⇒\frac{n+4}{n-4}=\frac{(n-4+8)}{(n-4)}=1+\frac{8}{n-4}\)
The expression is positive integer if \(\frac{8}{n-4}\) is integer.
Or \((n-4) \)must be a multiple of 8.

To maximize n, set \((n-4)\) equal to 8.

Therefore, n = 12.