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

Answer: (B)

12

Answer 2

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

$=\frac{n+4}{n-4}$

$⇒\frac{n+4}{n-4}=\frac{(n-4+8)}{(n-4)}=1+\frac{8}{n-4}$
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.