Question:
What is the largest positive integer n such that $\frac{n^{2}+7 n+12}{n^{2}-n-12}$ is also a positive integer?
- 8
- 12
- 16
- 6
Correct Answer: Option: 2
$\frac{{{n}^{2}}+7n+12}{{{n}^{2}}-n-12}=\frac{\left( n+3 \right)\left( n+4 \right)}{\left( n-4 \right)\left( n+3 \right)}=\frac{\left( n+4 \right)}{\left( n-4 \right)}$
$\Rightarrow \frac{\left( n+4 \right)}{\left( n-4 \right)}=\frac{\left( n-4+8 \right)}{\left( n-4 \right)}=1+\frac{8}{\left( n-4 \right)}$
The expression is positive integer if $\frac{8}{\left( n-4 \right)}$ is integer.
Or (n-4) must be factor of 8.
For n to be largest, n-4=8
Or n =12
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