Carcass wrote:
For which of the following values of n is \(\frac{96+n}{n}\) NOT an integer?
(A) 4
(B) 8
(C) 9
(D) 12
(E) 16
Key (and often-tested) property: \(\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}\)So, \(\frac{96+n}{n} = \frac{96}{n} + \frac{n}{n}= \frac{96}{n} + 1\)
Since 1 is an integer, the question boils down to finding a value of n so that \(\frac{96}{n}\) is NOT an integer
Since 96 is not the visible by 9, we know that \(\frac{96}{n}\) will not be an integer when n = 9
Answer: C