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

CAT 2019 Shift 2 Paper

Question : 100 of 100

Marks: +1, -0

What is the largest positive integers n such that

{n^2+7n+12}/{n^2-n-12}

is also a positive integers?

Solution:

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

⇒ \;{(n+4)}/{(n-4)}=\;{(n-4+8)}/{(n-4)}=1+\;{8}/{(n-4)}

The expression is positive integer if

\;{8}/{(n-4)}

is integer.
Or

( {n}-4)

must be factor of 8 .
For

n

to be largest,

n-4=8

{n}=12

Go to Question:

Prev Question